What value of n makes the equation 2/3 times in equals 4/9 true

Step-by-step explanation:

Check the attached file for solution and

simulation screen shot

R-Code:

Sample mean

sd = 60 Population Standard deviation

n = 10 Sample size

conf.level = 0.99 Confidence level

[tex]\alpha = 1-conf.level[/tex]

[tex]z\star = \round(\qnorm(1-\alpha/2),2); z.\star[/tex]

[tex]E = \round(z* \times \sigma/\sqrt(n),2); E[/tex]

[tex]x= c(E,-E)[/tex]

Answer:

Probability that it wins at least 3 of its final 5 games = .02387

Step-by-step explanation:

Given -

The probability of win the weekend game = 0.5

The probability of loose  the weekend game = 0.5

If he wins the game this weekend then it will play its final 5 games in the upper bracket of its league

In this case,  probability of success is (p) = 0.3

probability of failure is (q) = 1 - p = 0.7

Let X be number of game won out of last five games

probability that it wins at least 3 of its final 5 games

( 1 )

[tex]P(X\geq3)[/tex] = [tex]P(X\geq3/first\; game\; won)[/tex] ( probability of first game won )

               =   [tex]0.5\times[/tex]P( X =3 ) + [tex]0.5\times[/tex]P( X =4) + [tex]0.5\times P(X = 5)[/tex]

                =  [tex]0.5\times\binom{5}{3}(0.3)^{3}(0.7)^{2} + 0.5\times\binom{5}{4}(0.3)^{4}(0.7)^{1}[/tex] + [tex]0.5\times\binom{5}{5}(0.3)^{5}(0.7)^{0}[/tex]

                 = [tex]0.5\times\frac{5!}{(3!)(2!)}\times(0.3)^{3}\times(0.7)^{2} + 0.5\times\frac{5!}{(4!)(1!)}\times(0.3)^{4}\times(0.7)^{1}[/tex] + [tex]0.5\times\frac{5!}{(5!)(0!)}\times(0.3)^{5}\times(0.7)^{0}[/tex]= = .065 + .014 + .001215  = .080

If he loose the game this weekend then it will play its final 5 games in the lower bracket of its league

In this case,  probability of success is (s) = 0.4

probability of failure is (t) = 1 - s = 0.6

( 2 )

[tex]P(X\geq3/first\; game\; lost)[/tex] ( probability of first game lost )

= [tex]0.5\times P(X = 3) + 0.5\times P(X = 4)[/tex] + [tex]0.5\times P(X=5)[/tex]

= [tex]\binom{5}{3}(0.4)^{3}(0.6)^{2} + 0.5\times\binom{5}{4}(0.4)^{4}(0.6)^{1}[/tex]+ [tex]0.5\times\binom{5}{5}(0.4)^{5}(0.6)^{0}[/tex]

= [tex]0.5\times\frac{5!}{(3!)(2!)}\times(0.4)^{3}\times(0.6)^{2} + 0.5\times\frac{5!}{(4!)(1!)}\times(0.4)^{4}\times(0.6)^{1}[/tex] + [tex]0.5\times\frac{5!}{(5!)(0!)}\times(0.4)^{5}\times(0.6)^{0}[/tex] = = .1152 + .0384 + .00512 = .1587

Required probability = ( 1 ) + ( 2 ) = .02387

The restaurant pay $24,192 for 9 pounds of truffles.

The unitary technique involves first determining the value of a single unit, followed by the value of the necessary number of units.

For example, Let's say Ram spends 36 Rs. for a dozen (12) bananas.

12 bananas will set you back 36 Rs. 1 banana costs 36 x 12 = 3 Rupees.

As a result, one banana costs three rupees. Let's say we need to calculate the price of 15 bananas.

This may be done as follows: 15 bananas cost 3 rupees each; 15 units cost 45 rupees.

Given:

A restaurant buys 9 pounds of truffles.

as, 1 pound = 16 ounce

So, the restaurant buys

= 16 x 9

= 144 ounce of truffles.

If truffles cost $168 per ounce.

So, the cost for 9 pounds ruffles is

=  144 x 168

= $ 24,192

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Answer:

a) We need to conduct a hypothesis in order to test the claim that the true proportion p is greatr than 0.3, so then the system of hypothesis are.:  

Null hypothesis:[tex]p \leq 0.3[/tex]  

Alternative hypothesis:[tex]p > 0.3[/tex]  

Right tailed test

b) [tex]p_v =P(z>2.32)=0.0102[/tex]  

c) So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.1[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of interest is higher than 0.3

Step-by-step explanation:

Part a: Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion p is greatr than 0.3, so then the system of hypothesis are.:  

Null hypothesis:[tex]p \leq 0.3[/tex]  

Alternative hypothesis:[tex]p > 0.3[/tex]  

Right tailed test

When we conduct a proportion test we need to use the z statisitc, and the is given by:  

[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)  

The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].

Calculate the statistic  

For this case the statistic is given by [tex] z_{calc}= 2.32[/tex]

Part b: Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The next step would be calculate the p value for this test.  

Since is a right tailed test the p value would be:  

[tex]p_v =P(z>2.32)=0.0102[/tex]  

Part c

So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.1[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of interest is higher than 0.3

The hypothesis test is right-tailed. The P-value should be assessed using a standard normal distribution, and if it is less than the significance level of α0.10, the null hypothesis should be rejected. However, the exact P-value for z=2.32 needs to be determined before a decision can be made.

The test statistic of z=2.32 is obtained when testing the claim that p>0.3. This indicates the hypothesis test in question is right-tailed, as the alternative hypothesis (Ha) suggests that the proportion is greater than 0.3 (p>0.3).

To determine the P-value, we look at the area to the right of our z-test statistic in the standard normal distribution. Given that our z-value is 2.32, the P-value would typically be found using a z-table or statistical software. However, the provided reference states that for a z-test value of 3.32, which seems to be a typo since our z-value is 2.32, the P-value would be 0.0103. We need to correct this and find the P-value for z=2.32, which we would expect to be larger than the P-value for z=3.32 since 2.32 is closer to the mean of the standard normal distribution.

P-value interpretation is critical when deciding whether to reject the null hypothesis (H0). In this case, if we use a significance level of α=0.10, we compare the P-value to this significance level. If the P-value is less than α, we reject H0; if it's greater, we fail to reject H0. Without the exact P-value for z=2.32, we cannot make a definitive decision, but typically, a z-value of 2.32 would result in a P-value less than 0.10, which leads to rejection of H0.

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Answer: 8/10 or 4/5

Step-by-step explanation:

10/10 - 2/10 = 8/10

Answer:

Since 10 - 2 = 8

The fraction of the remaining oceans would be 8/10

And if you simplify both 8 and 10 by 2

Meaning you divide them by two

8 ÷ 2 = 4

10 ÷ 2 = 5

Our new fraction is 4/5

~DjMia~

The solution is, Volume of prism = 1/2x³ + x², is the expression which represents the volume of the prism, in cubic units.

Volume can be stated as the space taken by an object. Volume is a measure of three-dimensional space. Volume is defined as the space occupied within the boundaries of an object in three-dimensional space. It is also known as the capacity of the object.

here, we have,

Given that:

The oblique prism below has an isosceles right triangle base and the length of the base is x

=> the area of the base: 1/2 × x × x = 1/2

The vertical height of the prism is (x + 2)

=> The volume of the oblique prism is:

V = the base area * the vertical height

<=> V = 1/2* x² * (x + 2)

<=> V =  1/2x³ + x²

Hence, The solution is, Volume of prism = 1/2x³ + x², is the expression which represents the volume of the prism, in cubic units.

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Answer:

[tex]\Delta A = 5.9\,cm^{2}[/tex]

Step-by-step explanation:

The area of an rectangle is given by the following formula:

[tex]A = w\cdot h[/tex]

Where:

[tex]w[/tex] - Width, in centimeters.

[tex]h[/tex] - Height, in centimeters.

The differential of the expression is derived hereafter:

[tex]\Delta A = \frac{\partial A}{\partial w} \cdot \Delta w + \frac{\partial A}{\partial h}\cdot \Delta h[/tex]

[tex]\Delta A = h \cdot \Delta w + w \cdot \Delta h[/tex]

[tex]\Delta A = (31\,cm)\cdot (0.1\,cm) + (28\,cm)\cdot (0.1\,cm)[/tex]

[tex]\Delta A = 5.9\,cm^{2}[/tex]

Using differentials the maximum error in the calculated area of the rectangle wi’ould be 5.9 cm

The area formular of a rectangle is :

Maximum error can be defined thus :

Δmax = (L × Δe) + (W × Δe)

Δmax = (31 × 0.1) + (28 × 0.1)

Δmax = 3.1 + 2.8

Δmax = 5.9 cm

Hence, the maximum error in the calculated area value is 5.9 cm.

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Step-by-step explanation:

a) $1,000 or more?

As the data is usually distributed, using the 3-step approach to solve the problem: draw the image (I'm not going to do it here), locate the z-point, and identify the region on the graph.

(1000-951)/96 = .5104 = 0.51 = Z, so area = 1 -0.6950 = 0.3050

b) Between $900 and $1,100?

(900-951)/96 = -0.53 = Z, and (1100-951)/96 = 1.55 = Z So area = 0.9394 - 0.2981 = 0.6413

c) Between $825 and $925?

(825-951)/96 = -1.31 = Z, and (925-951)/96 = -0.27 = Z So area = 0.3936 - 0.0951 = 0.2985

d) Less than $700?

(700-951)/96 = -2.61 = Z, so area = 0.0045

The percentage increase in apartment supply is 30%, and the price sensitivity or price elasticity of supply is 3.90, indicating a relatively elastic supply to price changes.

The question deals with the concept of price elasticity of supply, which measures the responsiveness of the quantity supplied to a change in price. To determine the percentage increase in apartment supply, we calculate the change in quantity supplied, divide it by the original quantity supplied, and then multiply by 100 to convert to a percentage:

The original quantity supplied was 10,000 units, and the new quantity supplied is 13,000 units. So the change in quantity supplied is 13,000 - 10,000 = 3,000 units.

The percentage increase in supply is (3,000 units / 10,000 units) × 100% = 30%

Next, we calculate the price sensitivity, also known as price elasticity of supply, using the formula:

Price Elasticity of Supply (Es) = (% Change in Quantity Supplied) / (% Change in Price)

The price increased from $650 to $700, which is an increase of $700 - $650 = $50. The percentage change in price is ($50 / $650) × 100% = 7.69%

Then the price elasticity of supply is 30% / 7.69% = 3.90, indicating that the supply is relatively elastic.

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Answer:

##  Profit/Loss = [tex]14x-400[/tex]

##  29 tickets

Step-by-step explanation:

Profit/Loss is Revenue - Cost

For the fundraisers:

Revenue comes from tickets sold at $14 each

x tickets sold, means the revenue is:

14x

Now, cost is what they are going to give out, that is $400 TV Set, so the cost is:

400

Hence, Profit/Loss would be:

Profit/Loss = [tex]14x-400[/tex]

Raffle sales equaling the cost of prize is basically when we break-even, or when profit/loss is equal to 0. So we solve the equation:

Profit/Loss = 14x - 400

0 = 14x - 400

14x = 400

x = 28.57

We can't sell fractional tickets, so we have to sell 29 tickets in order to break even

Answer:

[tex]t=\frac{12.2-11.5}{\frac{1.6}{\sqrt{34}}}=2.551[/tex]    

[tex] df = n-1=34-1=33[/tex]

[tex]p_v =P(t_{(33)}>2.551)=0.008[/tex]  

Since the p value is less than the significance level of 0.05 we have enough evidence to reject the null hypothesis in favor of the claim

And the best conclusion for this case would be:

b)The p-value is 0.008, indicating sufficient evidence for his claim.

Step-by-step explanation:

Information provided

[tex]\bar X=12.2[/tex] represent the sample mean fould against

[tex]s=1.6[/tex] represent the sample standard deviation

[tex]n=34[/tex] sample size  

represent the value that we want to test

[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

[tex]p_v[/tex] represent the p value for the test (variable of interest)  

System of hypothesis

We need to conduct a hypothesis in order to check if the true mean is higher than 11.5 fouls per game:  

Null hypothesis:[tex]\mu \leq 11.5[/tex]  

Alternative hypothesis:[tex]\mu > 11.5[/tex]  

The statistic is given by:

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex]  (1)  

The statistic is given by:

[tex]t=\frac{12.2-11.5}{\frac{1.6}{\sqrt{34}}}=2.551[/tex]    

P value

The degreed of freedom are given by:

[tex] df = n-1=34-1=33[/tex]

Since is a one side test the p value would be:  

[tex]p_v =P(t_{(33)}>2.551)=0.008[/tex]  

Since the p value is less than the significance level of 0.05 we have enough evidence to reject the null hypothesis in favor of the claim

And the best conclusion for this case would be:

b)The p-value is 0.008, indicating sufficient evidence for his claim.

Answer:

  (-2, 3)

Step-by-step explanation:

In the form ...

  y -k = a(x -h)^2

the vertex is (h, k).

Your equation has k = 3, a = -1, h = -2, so the vertex is ...

  (h, k) = (-2, 3)

Answer:

7

Step-by-step explanation:

The surface area is the lateral area plus the base area. The base area is 9*5=45. Since there are two bases, its 90. Subtract that from 286 to get the lateral surface area, which is 196. The lateral surface area is the base perimeter*height. The base perimeter is 9+9+5+5= 28. 196/28=height. The height is 7

Answer:

Height= 7in

Step-by-step explanation:

Use the surface area formula and pulg in.

2(wl)(hl)(hw)=286

divide by 2 to get it to the other side.

143=45+9h+5h

subtract 45 and divide by 14 (9+5)

the height is 7

Answer:

So the Null hypothesis is rejected in this case

Step-by-step explanation:

  The number of celebrities is  n = 348

So to solve this we would assume that p is the percentage of people that died on the month preceding their birth month  

  Generally if there is no death postponement then p will be mathematically evaluated as

            [tex]p = \frac{1}{12}[/tex]  

This implies the probability of date in one month out of the 12 months

Now from the question we can deduce that the hypothesis we are going to be testing is  

  [tex]Null Hypothesis \ \ H_0 : p = 0.083[/tex]

This is a hypothesis is stating that a celebrity  dies in the month preceding their birth

  [tex]Alternative \ Hypothesis H_1 : p < 0.083[/tex]

   This is a hypothesis is stating that a celebrity does not die in the month preceding their birth

       is c is the represent probability for each celebrity which either c = 0 or c = 1

Where c = 0 is that the probability  that the celebrity does not die on the month preceding his/ her birth month

     and  c =  1  is that the probability  that the  celebrity dies on the month preceding his/ her birth month

  Then it implies that

   for  

       n= 1 + 2 + 3 + .... + 348  celebrities

Then the sum of c for each celebrity would be  [tex]c_s = 16[/tex]

i.e The number of celebrities that died in the month preceding their birth month

We are told that the significance level is  [tex]\alpha = 0.05[/tex], the the z value of [tex]\alpha[/tex] is

              [tex]z_{\alpha } = 1.65[/tex]

This is obtained from the z-table

Since this test is carried out on the left side of the area under the normal curve then the critical value will be

                 [tex]z_{\alpha } = - 1.65[/tex]

So what this implies is that  [tex]H_o[/tex] will be rejected if

                [tex]z \le -1.65[/tex]

Here z is the test statistics

Now z is mathematically evaluated as follows

                  [tex]z = \frac{c - np}{\sqrt{np_o(1- p_o)} }[/tex]

                [tex]z = \frac{16 - (348 *0.083)}{\sqrt{348*0.083 (1- 0.083)} }[/tex]

                [tex]z =-2.50[/tex]

From our calculation we see that the value of z is less than [tex]-1.65[/tex] so the Null hypothesis will be rejected

   Hence this tell us that the  evidence provided is not enough to conclude  that 16 celebrities died a month to their birth month

The question involves statistical hypothesis testing where the null hypothesis (H0) suggests no significant increase in celebrity deaths before their birth month, and the alternative hypothesis (H1) suggests a significant increase. Using significance level 0.05 and the provided data, the p-value is compared to decide on H0.

The question provided relates to setting up and testing a null hypothesis (H0) against a one-sided alternative hypothesis (H1) in the context of statistical hypothesis testing. Specifically, it involves determining whether the occurrence of celebrity deaths in the month preceding their birth month is statistically significant using a significance level of 0.05. To address this, the null hypothesis would state that there is no significant increase in the frequency of deaths in the month before the celebrities' birth month compared to any other month. The alternative hypothesis would state that there is a significant increase in deaths in the month preceding the birth month of celebrities. We would use the data provided (16 out of 348 celebrities dying in the month before their birth month) to calculate the p-value and compare it with the alpha level of 0.05 to decide whether to reject the null hypothesis or not.

The answer is c

Step-by-step explanation:

Answer:

answwr is c and i got it right

Step-by-step explanation:

Answer:

[tex]\Delta s \approx 754.579\,m[/tex] (See explanation below).

Step-by-step explanation:

Each floor has a height of 3 meters. Then, the number of floors of the cylinder is:

[tex]n = \frac{30\,m}{3\,m}[/tex]

[tex]n = 10\,floors[/tex]

Let consider that spiral makes a revolution per floor. Then, the parametric equations of the spiral are:

[tex]x = r\cdot \cos \theta[/tex]

[tex]y = r\cdot \sin \theta[/tex]

[tex]z = \Delta h \cdot \frac{\theta}{2\pi}[/tex]

Length of the staircase can be modelled by using the formula for arc length:

[tex]\Delta s = \int\limits^{20\pi}_{0} {\sqrt{\left(\frac{dx}{d\theta} \right) ^{2}+\left(\frac{dy}{d\theta} \right)^{2}+\left(\frac{dz}{d\theta}\right)^{2}}} \, d\theta[/tex]

[tex]\Delta s = \int\limits^{20\pi}_{0} {\sqrt{\left(-r\cdot \sin \theta\right)^{2}+\left(r\cdot \cos \theta\right)^{2}+\left(\frac{\Delta h}{2\pi} \right)^{2}} } \, d\theta[/tex]

[tex]\Delta s = \int\limits^{20\pi}_{0} {\sqrt{r^{2}+\frac{(\Delta h)^{2}}{4\pi^{2}} }} \, d\theta[/tex]

[tex]\Delta s = \sqrt{(12\,m)^{2}+\frac{(3\,m)^{2}}{4\pi^{2}} } \cdot (20\pi-0)[/tex]

[tex]\Delta s \approx 754.579\,m[/tex]

Answer:

base : 9

three points : (1,0), (9,1), (81,2)

domain : x>0

range : all real number

asymptote : x=0

Answer:

x=0

Step-by-step explanation:

I need to write a 5-paragraph eassy, so please help me it is base on an article name "Schools in Maryland Allow Elementary Students to Carry Cellphones, by Amanda Lenhart, The Washington Post" here are some pic. I just need help writing two paragraph. I already have my Introduction, Body Paragraph #1 and my Body paragraph #2 just need my Body paragraph #3 and my Conclusion I will give brainlis and 30 pnt.

Prompt:

Write an argumentative essay answering the questions: Should students be allowed to carry cellphones on campus? You must support your claim with evidence from the text. You may also use relevant examples from your own experience, observations, and other readings.

Directions:

Before you begin, read the text below, which presents information about the advantages and disadvantages of carrying a cell phone at school. Use the Student Writing Checklist on the back of this page to plan and write a multi-paragraph essay that addresses the prompt. Use your own words, except when quoting directly from the text.

PLEASE DONT WAST THEM

Answer:

60

Step-by-step explanation:

1/2 of 80 is 40.1/4 is 20.40 plus 20 equals 60.

Answer:

80x

Step-by-step explanation:

The probability is needed for this, so I put it as x.

If it was .25, he could expect to score 20 times.

Answer:

2%

Step-by-step explanation:

You do 12÷6×1=2

I used PEMDAS

Final answer:

The probability that Lue will roll a sum of 12 on two rolls of a standard six-sided die is 1/36 or about 2.78%, as only the combination (6,6) results in the sum of 12.

Explanation:

Probability of Rolling a Sum of 12

To calculate the probability that Lue will roll a sum of 12 on two rolls of a six-sided die, we need to consider all the possible combinations that can result in a sum of 12. These combinations are (6,6). Since each die is independent, we calculate the probability for one die and then square it for two dice, because there is only one way to get a six on a die, and there are six faces. Therefore, the probability of rolling a six is:

1/6

To find the probability of rolling two sixes, we multiply the probabilities of each independent event:

(1/6) × (1/6) = 1/36

So, the probability that Lue will roll a sum of 12 in two rolls is 1/36, or approximately 2.78%.

Answer:

z statistic = 2.82

Step-by-step explanation:

Sample size = 1200

Null hypothesis, [tex]H_{0}[/tex]: p = 0.58

Alternative hypothesis, [tex]H_{a}[/tex]: p > 0.58

From the null and alternative hypothesis, we can derive that Hypothesized proportion, [tex]p_{0}[/tex] = 0.58 = 58%

Significance level = 2.5% = 0.025

Sample proportion, [tex]p_{1}[/tex] = 0.62 = 62%

Test statistic, z:

[tex]z_{statistic} = \frac{p_{1} -p_{0} }{\sqrt{\frac{p_{0}( 1 -p_{0} )}{n} } }[/tex]

[tex]z_{statistic} = \frac{0.62-0.58}{\sqrt{\frac{0.58(1-0.58)}{1200} } }[/tex]

[tex]z statistic =\frac{0.04}{\sqrt{0.000203} }[/tex] [tex]= \frac{0.04}{0.0142}[/tex]

[tex]z statistic = 2.82[/tex]

The value of the test statistic 'z' for the given hypothesis test is approximately 5.65, calculated using the provided sample proportion, hypothesized population proportion, and sample size.

The question asks for the calculation of the test statistic 'z' for a hypothesis test where the null hypothesis states that the population proportion is equal to .58 and the alternative hypothesis says that the population proportion is greater than .58. This is tested at a 2.5% significance level with a sample size of 1200 and sample proportion of .62.

To calculate the test statistic z, use the formula:

z = (p' - p) / √(p(1 - p)/n)

where:

Now, plug in the values:

z = (0.62 - 0.58) / √(0.58(1 - 0.58)/1200)

z ≈ 5.65

This is the value of the test statistic z for conducting the hypothesis test.

Answer:

a. (4, 650), (8, 400)

b. -62.5

Step-by-step explanation:

a......................

Data points are (4, 650) and (8, 400) since price is independent variable and weight of coffee is dependent

b......................

Slope = (y2-y1)/(x2-x1)

Slope = (400 - 650)/(8-4) = -250/4 = - 62.5

The data points representing the price and quantity of coffee sold are ($4, 650) and ($8, 400). The slope of the line connecting these points, which shows the relationship between price and quantity, is -62.5.

This question is about mathematical calculations used in economics, specifically related to pricing and demand. Let's turn these scenarios into data points, with price ($4 and $8) as the independent variable and quantity (650 lbs and 400 lbs) as the dependent variable. So, your data points are ($4, 650) and ($8, 400).

To find the slope of the line connecting these points, use the slope formula. The slope(m) is equal to the difference in the y-values divided by the difference in the x-values. So, m = (400-650) / (8-4) = -250/4 = -62.5.

Thus, the slope of the line connecting these two points, representing the relationship between the price and quantity of coffee sold, is -62.5.

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Answer:

80 meters (8*10=80)

Answer:

80

Step-by-step explanation:

10 times 8= 80

to find the area is always lenght × height × weight

to find the perimeter is always lenght × lenght × heigth × heigth

example...

a house with the height of 5 and the lenght of 1 .find the perimeter

5+5+1+1= 12

Answer:0.6 sec

Step-by-step explanation:

Answer:

Step-by-step explanation:

0.6 is the answer just took the test

Answer:

Step-by-step explanation:

He invested $41,000 more than 4 times the amount at 9%.

Total Annual Interest Earned = $35,260

Therefore, Time=1 year

Simple Interest[tex]=\frac{Principal X Rate X Time}{100}[/tex]

Therefore, his total interest

=Interest from Investment 1 + Interest from Investment 2

[tex]35260=\left(\frac{x*5*1}{100} \right)+\left(\frac{4x+41000*9*1}{100} \right)\\35260=0.05x+(0.36x+3690)\\35260-3690=0.05x+0.36x\\31570=0.41x\\\text{Divide both sides by 0.41}\\x=\$77000[/tex]

Therefore:

The amount invested at 5%=$77,000

The amount invested at 9%=$(4*77,000+41000)=$349,000

When you have a negative exponent, you move the base with the negative exponent to the other side of the fraction to make the exponent positive.

For example:

[tex]\frac{1}{2y^{-3}} =\frac{y^3}{2}[/tex]    ("y" is the base with the negative exponent)

[tex]x^{-5}[/tex] or [tex]\frac{x^{-5}}{1} =\frac{1}{x^5}[/tex]

When you multiply an exponent directly to a base with an exponent, you multiply the exponents together.

For example:

[tex](y^3)^2=y^{(3*2)}=y^6[/tex]

[tex](x^2)^4=x^{(2*4)}=x^8[/tex]

[tex](2n)^3[/tex] or [tex](2^1n^1)^3=2^{(1*3)}n^{(1*3)}=2^3n^3=8n^3[/tex]

When you have an exponent of 0, the result will always equal 1

For example:

[tex]x^0=1[/tex]

[tex]5^0=1[/tex]

[tex]y^0=1[/tex]

[tex](6m^{-1}*n^0)^{-3}[/tex]      I think you should first make the exponents positive

[tex]\frac{1}{(\frac{6}{m^1} *n^0)^3}[/tex]    

Since you know:

m = 3

n = -5    Substitute/plug it into the equation

[tex]\frac{1}{(\frac{6}{(3)^1}*(-5)^0)^3 }[/tex]

[tex]\frac{1}{(2*1)^3}[/tex]

[tex]\frac{1}{2^3}[/tex]

[tex]\frac{1}{8}[/tex]      

Answer:

0.1667

Step-by-step explanation:

There are 6! ways to arrange the hats. The number of ways for which no guest will receive the proper hat is 5! (since there are 5 wrong hats for the first guest, 4 for the second guest, and so on). The probability that no guest will receive the proper hat is:

[tex]P=\frac{5!}{6!}=0.1667[/tex]

The probability is 0.1667.

Answer:

[tex](x, y) = \left(\frac{3}{2}, \frac{\sqrt{3}}{2}\right)[/tex]

Step-by-step explanation:

The rectangular coordinates of the point are:

[tex](x,y) = \left(\sqrt{3}\cdot \cos\frac{\pi}{6}, \sqrt{3}\cdot \sin\frac{\pi}{6}\right)[/tex]

[tex](x, y) = \left(\frac{3}{2}, \frac{\sqrt{3}}{2}\right)[/tex]

Answer:

B

Step-by-step explanation:

At a 5 percent significance level and with a p-value of 0.023, we reject the null hypothesis, concluding that more than 20% of cars are speeding.

The question involves determining whether to reject the null hypothesis based on a p-value from a statistical test concerning the true proportion of cars that are speeding on a highway. Since the p-value of 0.023 is less than the alpha level of 0.05, we would reject the null hypothesis (H0: p = 0.20). At the 5 percent significance level, there is sufficient evidence to conclude that more than 20% of cars are speeding on the highway.

Since the p-value is less than the alpha level of 0.05, we reject the null hypothesis. Therefore, the police department should consider installing the speed sensor and traffic camera.

To determine if the police department should install a speed sensor and traffic camera based on a significance test, we need to examine the hypotheses:

H0: p = 0.20 (the true proportion of cars speeding is 20%)

H1: p > 0.20 (the true proportion of cars speeding is greater than 20%)

Given that in a random sample of 100 cars, 28 were speeding, the test resulted in a p-value of 0.023. At the alpha level of 0.05, since the p-value (0.023) is less than alpha (0.05), we reject the null hypothesis.

In conclusion, at the 5 percent significance level, there is sufficient evidence to conclude that the true proportion of cars speeding is greater than 20%, justifying the installation of the speed sensor and traffic camera.

Answer:

1. y + 10 - 3/2y = -y/2 + 10

2. 2r+ 7r-r - 9 = 8r - 9

3.  7 + 4p-5+p+2q = 2 + 5p + 2q

Step-by-step explanation:

basically you can add terms that have the same variable

integers can be added together, Xs can be added, Zs, Ys, As, Bs, Cs, you get the point

1. y + 10 - 3/2y = -y/2 + 10

2. 2r+ 7r-r - 9 = 8r - 9

3.  7 + 4p-5+p+2q = 2 + 5p + 2q (do not add different variables p and q ) together

try 4-6 on your own to get this skill down, if you need help with those just let me know

Answer:

For the maximize the total area, the all wire i.e. 26 m should be used for the square.

Step-by-step explanation:

Length of the wire = 26 m

Let Amount of wire cut for square = x

Amount of wire cut for triangle  = 26 - x

Side of the square = [tex]\frac{x}{4}[/tex]

Area of the square = [tex]\frac{x^{2} }{16}[/tex]  ------ (1)

Side of the triangle is given by

[tex]a = \frac{26 - x}{3}[/tex]

Side of the triangle is [tex]a = \frac{26 - x}{3}[/tex]

Area of the triangle is given by

[tex]A = \frac{\sqrt{3} }{4} a^{2}[/tex]

Area of the triangle is

[tex]A =\frac{\sqrt{3} }{36} (26 - x)^{2}[/tex]  ------- (2)

Now the total area = Area of square + Area of triangle

The total area = [tex]\frac{x^{2} }{16} + \frac{\sqrt{3} }{36} (26 - x)^{2}[/tex]  ------- (3)

Differentiate above equation with respect to x we get

[tex]A' = \frac{x}{8} - \frac{\sqrt{3} }{18} (26 - x)[/tex]

Take [tex]A' = 0[/tex]

[tex]\frac{x}{8} - \frac{\sqrt{3} }{18} (26 - x) = 0[/tex] ------- (4)

By solving the above equation we get

x = 11.31 m

Again take [tex]A''[/tex] by differentiating equation  (4)

[tex]\frac{x}{8} + \frac{\sqrt{3} }{18} (26)[/tex]

Which is greater than zero. so the value x = 11.31 m gives the area minimum.

Thus for the maximize the total area, the all wire i.e. 26 m should be used for the square.

Answer:

The probability that at least two-third of vehicles in the sample turn is 0.4207.

Step-by-step explanation:

Let X = number of vehicles that turn left or right.

The proportion of the vehicles that turn is, p = 2/3.

The nest n = 50 vehicles entering this intersection from the east, is observed.

Any vehicle taking a turn is independent of others.

The random variable X follows a Binomial distribution with parameters n = 50 and p = 2/3.

But the sample selected is too large and the probability of success is close to 0.50.

So a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:

Check the conditions as follows:

[tex]np=50\times \frac{2}{3}=33.333>10\\\\n(1-p)=50\times \frac{1}{3}= = 16.667>10[/tex]

Thus, a Normal approximation to binomial can be applied.

So, [tex]X\sim N(np, np(1-p))[/tex]

Compute the probability that at least two-third of vehicles in the sample turn as follows:

[tex]P(X\geq \frac{2}{3}\times 50)=P(X\geq 33.333)=P(X\geq 34)[/tex]

                        [tex]=P(\frac{X-\mu}{\sigma}>\frac{34-33.333}{\sqrt{50\times \frac{2}{3}\times\frac {1}{3}}})[/tex]

                        [tex]=P(Z>0.20)\\=1-P(Z<0.20)\\=1-0.5793\\=0.4207[/tex]

Thus, the probability that at least two-third of vehicles in the sample turn is 0.4207.