2x²(3x²+7x-6)
Can anyone solve it​

Answer: No, these results are not statistically significant because

p > 0.05

Step-by-step explanation:

The null hypothesis is

H0 : μ = 17

The alternative hypothesis is

H 1 : μ ≠ 17

where μ is the mean amount of cereal in each box.

The p value that he got is 0.1499. This is greater than alpha = 0.05 which is the given level of significance.

If the level of significance is lesser than the p value, we would accept the null hypothesis.

Therefore, the correct option is

No, these results are not statistically significant because p>0.05

Answer:

Yes, this provide enough evidence to show a difference in the proportion of households that own a​ desktop.

Step-by-step explanation:

We are given that National data indicates that​ 35% of households own a desktop computer.

In a random sample of 570​ households, 40% owned a desktop computer.

Let p = population proportion of households who own a desktop computer

SO, Null Hypothesis, [tex]H_0[/tex] : p = 25%   {means that 35% of households own a desktop computer}

Alternate Hypothesis, [tex]H_A[/tex] : p [tex]\neq[/tex] 25%   {means that % of households who own a desktop computer is different from 35%}

The test statistics that will be used here is One-sample z proportion statistics;

                                  T.S.  = [tex]\frac{\hat p-p}{{\sqrt{\frac{\hat p(1-\hat p)}{n} } } } }[/tex]  ~ N(0,1)

where, [tex]\hat p[/tex] = sample proportion of 570​ households who owned a desktop computer = 40%

            n = sample of households = 570

So, test statistics  =  [tex]\frac{0.40-0.35}{{\sqrt{\frac{0.40(1-0.40)}{570} } } } }[/tex]

                               =  2.437

Since, in the question we are not given with the level of significance at which to test out hypothesis so we assume it to be 5%. Now at 5% significance level, the z table gives critical values of -1.96 and 1.96 for two-tailed test. Since our test statistics doesn't lies within the range of critical values of z so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that % of households who own a desktop computer is different from 35%.

Answer:

[tex]Z = -2.865[/tex] means that it would be unusual to see 32% who rarely or never use academic advising services in one of these samples

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Z scores below -2 are considered unusually low.

Z scores above 2 are considered unusually high.

For a sample proportion p in a sample of size n, we have that [tex]\mu = p, \sigma = \sqrt{\frac{p(1-p)}{n}}[/tex]

In this problem, we have that:

[tex]\mu = 0.42, \sigma = \sqrt{\frac{0.42*0.58}{200}} = 0.0349[/tex]

Is it unusual to see 32% who rarely or never use academic advising services in one of these samples

What is the z-score for X = 0.32?

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{0.32 - 0.42}{0.0349}[/tex]

[tex]Z = -2.865[/tex]

[tex]Z = -2.865[/tex] means that it would be unusual to see 32% who rarely or never use academic advising services in one of these samples

No, it is not unusual to see 32% of students rarely or never using academic advising services in one of these samples.

To determine if it is unusual to see 32% of students rarely or never using academic advising services in one of these samples, we can compare it to the range of values that would be considered usual. In this case, we can use the 95% confidence interval provided, which states that the true proportion of community college students who rarely or never use academic advising services is between 0.113 and 0.439. If the observed proportion falls within this interval, it would be considered usual; otherwise, it would be considered unusual.

Since 32% falls within the range of 0.113 and 0.439, it is considered a usual value. Therefore, it is not unusual to see 32% of students rarely or never use academic advising services in one of these samples.

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Answer:

There is no relationship between your year in school and having a job.

Step-by-step explanation:

In this instance, the chi sq test need to be performed.

Chi sq is used to determine if there is a significant relationship between two categorical variables.

The two variables here are year in school and employment status.

The two variables are independent(no relationship exists)

This implies that there is null hypothesis

Therefore, the Null Hypothesis is

There is no relationship between your year in school and having a job.

Answer:

You will need 10 boxes

Step-by-step explanation:

Nine boxes will only hold 54 trophies so you need one more than nine.

Answer:

You will need 348 boxes for the trophies

Step-by-step explanation:

just multiply 58 and 6

The correct question is:

Write out the following sums, one term for each value of k. Simplify each term as much as possible, but do not enter decimals. For example, enter 1 + 4 + 9 instead of 1² + 2² + 3² or 14, or enter 1/2 + 1/2 instead of 0.5 + 0.5 or 1.

The purpose of this problem is for you to show that you know how to interpret summation notation and write all of the terms in a sum, which is why you are being told not to reduce your answers very much.

[tex](a) \sum_{k=0}^5 2^k \\ \\(b) \sum_{k=2}^7 \frac{1}{k} \\ \\(c) \sum_{k=1}^5 k^2 \\ \\(d) \sum_{k=1}^6 \frac{1}{6} \\ \\(e) \sum_{k=1}^6 2k[/tex]

Answer:

[tex](a) \sum_{k=0}^5 2^k = $1 + 2 + 4 + 8 + 16 + 32$ \\ \\(b) \sum_{k=2}^7 \frac{1}{k} = \frac{1}{2} + \frac{1}{3} + \frac{1}{4}+ \frac{1}{5}+ \frac{1}{6}+ \frac{1}{7} \\ \\(c) \sum_{k=1}^5 k^2 = 1 + 4 + 9 + 16 + 25 \\ \\(d) \sum_{k=1}^6 \frac{1}{6} = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} \\ \\(e) \sum_{k=1}^6 2k = 2 +4 +6 +8 +10 +12[/tex]

Step-by-step explanation:

[tex](a) \sum_{k=0}^5 2^k\\For k = 0: 2^k = 2^0 = 1\\For k = 1: 2^1 = 2\\For k = 2: 2^2 = 4\\For k = 3: 2^3 = 8\\For k = 4: 2^4 = 16\\For k = 5: 2^5 = 32\\\sum_{k=0}^5 2^k = 1 + 2 + 4 + 8 + 16 + 32[/tex]

[tex](b) \sum_{k=2}^7 \frac{1}{k}\\For k = 2: 1/2\\For k = 3: 1/3\\For k = 4: 1/4\\For k = 5: 1/5\\For k = 6: 1/6\\For k = 7: 1/7\\ \sum_{k=2}^7 \frac{1}{k} = 1/2 + 1/3 + 1/4 + 1/5 + 1/6+ 1/7[/tex]

[tex](c) \sum_{k=1}^5 k^2\\For k = 1: 1^2 = 1\\For k = 2: 2^2 = 4\\For k = 3: 3^2 = 9\\For k = 4: 4^2= 16\\For k = 5: 5^2 = 25\\\sum_{k=1}^5 k^2 = 1 + 4 + 9 + 16 + 25[/tex]

[tex](d) \sum_{k=1}^6 \frac{1}{6}\\For k = 1: 1/6\\For k = 2: 1/6\\For k = 3: 1/6\\For k = 4: 1/6\\For k = 5: 1/6\\For k = 6: 1/6\\ \sum_{k=1}^6 \frac{1}{6} = 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6[/tex]

[tex](e) \sum_{k=1}^6 2k\\For k = 1: 2\times1 = 2\\For k = 2: 2\times2 = 4\\For k = 3: 2\times3 = 6\\For k = 4: 2\times4 = 8\\For k = 5: 2\times5 = 10\\For k = 6: 2\times6 = 12\\\sum_{k=1}^6 2k = 2 +4 +6 +8 +10 +12[/tex]

Answer:

c

Step-by-step explanation:

Answer:

12.75

Step-by-step explanation:

Answer:

x=4 y=5

Step-by-step explanation:

5x + 2y = 30 and -5x + 4y = 0

Solve by adding the two equations together to eliminate x

5x + 2y = 30

-5x + 4y = 0

----------------------

   6y = 30

Divide each side by 6

6y/6 = 30/6

y =5

Now we solve for x

5x + 2y = 30

5x +2(5) = 30

5x+10 = 30

Subtract 10 from each side

5x+10-10 = 30-10

5x = 20

Divide each side by 5

5x/5 = 20/5

x = 4

The amount the florist earns for each delivery, with 'x' being the miles away from the flower shop, can be modeled with the equation: Earnings = (7.5 + 0.75x). This represents the fixed net income of $7.5 and $0.75 per mile after paying the driver.

The florist charges ​$12.00 for delivery and an additional ​$1.50 per mile from the flower shop. However, the florist also has costs to cover, namely $0.75 per mile to pay the driver, and ​$4.50 for gas per delivery. The net earning per delivery, with 'x' representing the number of miles a delivery location is from the flower shop, can be modeled by the following algebraic expression: Earnings = (12 + 1.5x) - (0.75x + 4.5).

This actually simplifies to: Earnings = (7.5 + 0.75x). The 7.5 is the fixed net income for each delivery (gross earnings minus the gasoline cost) and 0.75x is the per-mile net income after the driver is paid.

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The original price of the guitar was $1250, calculated by taking the sale price of $825 and dividing it by 0.66, which represents the remaining percentage of the price after a 34% discount.

To find the original price of the guitar that has been marked down by 34% and sold for $825, we first need to consider that after the markdown, the guitar's price is equivalent to 100% - markdown percentage of the original price. In this case, it's 100% - 34% = 66% of the original price.

Let's denote the original price by P. Since 66% of P equates to $825, we can set up the following equation:

0.66  imes P = $825

Solving for P:

P = $825 / 0.66

P = $1250

Therefore, the original price of the guitar was $1250.

Answer:

The different ways in which Keely can pick the food and drinks to serve at the bridal shower is 2,772,000.

Step-by-step explanation:

Combinations is a mathematical procedure to determine the number of ways to select k items from n different items, without replacement and irrespective of the order of selection.

The formula to compute the combination of k items from n items is:

[tex]{n\choose k}=\frac{n!}{k!(n-k)!}[/tex]

The menu for the bridal shower consists of:

Beverages: 4

Appetizers: 3

Dessert: 4

It is provided that Keely does not have time to cook. SO she goes to the supermarket and there she has the following options:

Beverages: 11

Appetizers: 10

Dessert: 8

Compute the number of ways Keely can select 4 beverages from 11 bottled drinks as follows:

[tex]{n\choose k}=\frac{n!}{k!(n-k)!}[/tex]

[tex]{11\choose 4}=\frac{11!}{4!(11-4)!}[/tex]

      [tex]=\frac{11!}{4!\times 7!}[/tex]

      [tex]=\frac{11\times 10\times 9\times 8\times 7!}{4!\times 7!}[/tex]

      [tex]=330[/tex]

Keely can select 4 beverages in 330 ways.

Compute the number of ways Keely can select 3 appetizers from 10 frozen appetizers as follows:

[tex]{n\choose k}=\frac{n!}{k!(n-k)!}[/tex]

[tex]{10\choose 3}=\frac{10!}{3!(10-3)!}[/tex]

      [tex]=\frac{10!}{3!\times 7!}[/tex]

      [tex]=\frac{10\times 9\times 8\times 7!}{3!\times 7!}[/tex]

      [tex]=120[/tex]

Keely can select 3 appetizers in 120 ways.

Compute the number of ways Keely can select 4 desserts from 8 prepared desserts as follows:

 [tex]{n\choose k}=\frac{n!}{k!(n-k)!}[/tex]

 [tex]{8\choose 4}=\frac{8!}{4!(8-4)!}[/tex]

      [tex]=\frac{8!}{4!\times 4!}[/tex]

      [tex]=\frac{8\times 7\times 6\times 5\times 4!}{4!\times 4!}[/tex]

      [tex]=70[/tex]

Keely can select 4 desserts in 70 ways.

Compute the total number of ways in which Keely can select 4 beverages, 3 appetizers, and 4 desserts for the party as follows:

Total number of ways = n (4 beverages) × n (appetizers) × n (dessert)

                                    [tex]={11\choose 4}\times {10\choose 3}\times {8\choose 4}[/tex]

                                    [tex]=330\times 120\times 70\\=2772000[/tex]

Thus, the different ways in which Keely  can pick the food and drinks to serve at the bridal shower is 2,772,000.

Answer:

a) 50

b) 6.71

c) 0.0681

Step-by-step explanation:

check the attached file below

Solving for the velocity v(t) of a cannonball considering air resistance involves integrating the differential equation m dv/dt = -mg - kv, where k is a constant of proportionality. Given initial conditions, this allows one to calculate the cannonball's velocity at any time.

A student has asked how to find the velocity v(t) of a cannonball, considering air resistance, modeled by m dv/dt = -mg - kv, where m is the mass of the cannonball, g is the acceleration due to gravity (32 ft/s2), and k is a constant of proportionality (0.0025). Given the initial velocity v0 = 290 ft/s, the solution involves solving this differential equation with the given initial condition.

However, solving this specific differential equation requires integration techniques that account for the linear dependence of the air resistance on the velocity, which will yield an expression for v(t) as a function of time t. This formula can then be used to calculate the velocity of the cannonball at any given time, illustrating how air resistance affects its ascent and eventual descent.

Allocating 3 portions to x and 7 portions to y.

Step-by-step explanation:

Given that,

The percentage of students buy lunch at cafeteria: 30% = x = 0.3

Hence, the students bringing the lunch from home would be = y = 1 - 0.3 = 0.7

Now, the spinner has that equal-sized sections. Also, the probability of x and y are 0.3 and 0.7.

After multiplying both the probability by 10, we get

x = 3

y = 7

It shows that for every three students who buy lunch from cafeteria, seven students bring food from home. Hence, we can allocate the side of spinner for simulation in such as way:

Section 0 = y

Section 1 = y

Section 2 = x

Section 3 = y

Section 4 = y

Section 5 = x

Section 6 = y

Section 7 = y

Section 8 = x

Section 9 = y

Let's try to complete the squares.

The x-part starts with [tex]x^2+10x[/tex], which is the beginning of [tex]x^2+10x+25=(x+5)^2[/tex]. So, we'll think of [tex]x^2+10x[/tex] as [tex](x+5)^2-25[/tex]

Similarly, we have that

[tex]y^2+12y = (y+6)^2-36[/tex]

So, the equation becomes

[tex]x^2 + y^2 + 10x + 12y + 25 = 0 \iff (x+5)^2-25 + (y+6)^2-36+25=0 \iff (x+5)^2+ (y+6)^2-36=0 \iff (x+5)^2+ (y+6)^2=36[/tex]

Now we have writte the equation of the circle in the form

[tex](x-k)^2+(y-h)^2=r^2[/tex]

When the equation is in this form, everything is more simple: the center is [tex](k,h)[/tex] and the radius is [tex]r[/tex].

Answer:

Center// (-5,-6)

Radius// 6

Answer:

2.28% probability that the mean amount of their monthly cell phone bills is more than $60

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

[tex]\mu = 58.90, \sigma = 3.64, n = 44, s = \frac{3.64}{\sqrt{44}} = 0.54875[/tex]

What is the probability that the mean amount of their monthly cell phone bills is more than $60?

This is 1 subtracted by the pvalue of Z when X = 60. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{60 - 58.90}{0.54875}[/tex]

[tex]Z = 2[/tex]

[tex]Z = 2[/tex] has a pvalue of 0.9772

1 - 0.9772 = 0.0228

2.28% probability that the mean amount of their monthly cell phone bills is more than $60

Equation used to determine the surface area of the box : 2( lb + bh + lh)

Surface area of the box : 94in²

Given, A box is 4 inches wide, 5 inches long and 3 inches tall.

Formula of surface area of cuboid : 2( lb + bh + lh)

Here,

l = 5in

b = 4in

h = 3in

Substitute the values,

Surface area = 2(5×4 + 4×3 + 5×3)

Surface area = 94in²

Know more about surface area,

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Answer:

it would be 5 feet

Step-by-step explanation:

Answer:

[tex] \overline{AB} \cong \overline{DB} \:\: and \:\: \overline{AC} \cong \overline{DC}[/tex]

Step-by-step explanation:

[tex] \red{ \boxed{ \bold{\overline{AB} \cong \overline{DB} \:\: and \:\: \overline{AC} \cong \overline{DC}}} }\\ \overline{BC} \cong \overline{BC}... (common\: side) \\ [/tex]

Answer:

The critical value of t for 99% confidence interval is 2.898.

Step-by-step explanation:

The complete question is:

A biologist is trying to determine the average age of a local forest. She cuts down 18 randomly  selected trees and counts the tree rings. They find the average number of tree rings to be 83 with a  variance of 320. What is the critical value for the 99% confidence interval?

The population variance is not known and the sample size is too small. So a t-confidence interval will be used to estimate the population mean age of a local forest.

The (1 - α)% confidence interval for population mean is:

[tex]CI=\bar x\pm t_{\alpha/2, (n-1)}\times \frac{s}{\sqrt{n}}[/tex]

The information provided is:

n = 18

(1 - α)% = 99%

The degrees of freedom of the critical value of t is:

n - 1 = 18 - 1 = 17

Compute the critical value of t as follows:

[tex]t_{\alpha/2, (n-1)}=t_{0.01/2, (18-1)}=t_{0.005, 17}=2.898[/tex]

*Use a t-table.

Thus, the critical value of t for 99% confidence interval is 2.898.

Answer:

Dr. Potter gave 32 polio vaccinations and 28 measles vaccinations

Step-by-step explanation:

Total of 184 doses

Polio vaccination= 4 doses

Measles vaccination=2 doses

184=4p+2m

92=2p+m

Lets plug in 32+28

92=(2*32)+28

92=64+28

p=32, m=28

Answer:

The average speed was 48.7 miles per hour.

Step-by-step explanation:

The average speed v is given by the following formula:

[tex]v = \frac{d}{t}[/tex]

In which d is the distance, and t is the time.

The Harrisons drove 304.2 miles in 6.25 hours

This means that [tex]d = 304.2, t = 6.25[/tex]

We have the distance is miles and the time in hours, so the distance is in miles per hour.

So

[tex]v = \frac{304.2}{6.25} = 48.7[/tex]

The average speed was 48.7 miles per hour.

Answer:

[tex](\, \cos(\frac{\pi}{16}) + i\sin(\frac{\pi}{16}) \,)^{1/2} = \cos(\frac{\pi}{32}) + i\sin(\frac{\pi}{32}) = 0.99 + i0.09[/tex]

Step-by-step explanation:

The complex number given is

[tex]z = (\, \cos(\frac{\pi}{16}) + i\sin(\frac{\pi}{16}) \,)^{1/2}[/tex]

Now, remember that the DeMoivre's theorem states that

[tex]( \cos(x) + i\sin(x) )^n = \cos(nx) + i\sin(nx)[/tex]

Then for this case we have that

[tex](\, \cos(\frac{\pi}{16}) + i\sin(\frac{\pi}{16}) \,)^{1/2} = \cos(\frac{\pi}{32}) + i\sin(\frac{\pi}{32}) = 0.99 + i0.09[/tex]

Answer:

[tex]z=\frac{0.095-0.125}{\sqrt{0.11(1-0.11)(\frac{1}{200}+\frac{1}{200})}}=-0.959[/tex]    

[tex]p_v =P(Z<-0.959)=0.169[/tex]    

Comparing the p value with the significance level assumed[tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to to FAIL to reject the null hypothesis, and we can't conclude that company A are more reliable than televisions from company B at 5% of significance.

Step-by-step explanation:

Data given and notation    

[tex]X_{1}=19[/tex] represent the number of tvs who need a repair for A

[tex]X_{2}=25[/tex] represent the number of tvs who need a repair for B

[tex]n_{1}=200[/tex] sample 1 selected  

[tex]n_{2}=200[/tex] sample 2 selected  

[tex]p_{1}=\frac{19}{200}=0.095[/tex] represent the proportion estimated for the sample A  

[tex]p_{2}=\frac{25}{200}=0.125[/tex] represent the proportion estimated for the sample B

[tex]\hat p[/tex] represent the pooled estimate of p

z would represent the statistic (variable of interest)    

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

[tex]\alpha=0.05[/tex] significance level given  

Concepts and formulas to use    

We need to conduct a hypothesis in order to check if company A are more reliable than televisions from company B (that means p1<p2) , the system of hypothesis would be:    

Null hypothesis:[tex]p_{1} \geq p_{2}[/tex]    

Alternative hypothesis:[tex]p_{1} < p_{2}[/tex]    

We need to apply a z test to compare proportions, and the statistic is given by:    

[tex]z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}[/tex]   (1)  

Where [tex]\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{19+25}{200+200}=0.11[/tex]  

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.    

Calculate the statistic  

Replacing in formula (1) the values obtained we got this:    

[tex]z=\frac{0.095-0.125}{\sqrt{0.11(1-0.11)(\frac{1}{200}+\frac{1}{200})}}=-0.959[/tex]    

Statistical decision  

Since is a left sided test the p value would be:    

[tex]p_v =P(Z<-0.959)=0.169[/tex]    

Comparing the p value with the significance level assumed[tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to to FAIL to reject the null hypothesis, and we can't conclude that company A are more reliable than televisions from company B at 5% of significance.    

Final answer:

To determine if televisions from company A are more reliable than televisions from company B, we can perform a hypothesis test.

Explanation:

To determine if televisions from company A are more reliable than televisions from company B, we can perform a hypothesis test. We will compare the proportions of televisions requiring repairs in the two companies.

Step 1: State the hypotheses:

H0: The proportion of televisions requiring repairs in company A is the same as in company B.

HA: The proportion of televisions requiring repairs in company A is less than in company B.

Step 2: Set the significance level, let's say α = 0.01.

Step 3: Calculate the test statistic. We will use the Z-test for comparing proportions.

Step 4: Calculate the p-value.

Step 5: Compare the p-value to the significance level. If the p-value is less than α, we reject the null hypothesis and conclude that televisions from company A are more reliable than televisions from company B.

By performing the above steps, we can determine if the data shows that televisions from company A are more reliable than televisions from company B.

Answer:

0% probability you are guilty

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

[tex]\mu = 1, \sigma = 0.02[/tex]

If the metal is deemed faulty when the depth of penetration is more than 1.3 millimeters, what is the probability you are guilty?

This is 1 subtracted by the pvalue of Z when X = 1.3. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{1.3 - 1}{0.02}[/tex]

[tex]Z = 15[/tex]

[tex]Z = 15[/tex] has a pvalue of 1

1 - 1 = 0

0% probability you are guilty

We can use the z-score formula to calculate the probability of being guilty of distributing faulty metal based on the depth of penetration. The probability is practically zero.

To find the probability that you are guilty of distributing faulty metal, we need to calculate the probability that the depth of penetration is more than 1.3 millimeters. Since the depth of penetration is normally distributed with a mean of 1 millimeter and a standard deviation of 0.02 millimeters, we can use the z-score formula to standardize the value. The z-score is calculated as (x - μ) / σ, where x is the value we want to standardize, μ is the mean, and σ is the standard deviation.

Substituting the values into the formula, we have z = (1.3 - 1) / 0.02 = 15. Therefore, we need to find the probability that the z-score is greater than 15. Using a standard normal distribution table or calculator, we find that this probability is practically zero. Hence, the probability that you are guilty is practically zero.

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The volume of the rectangular prism with dimensions 10/3 cm, 4/5 cm, and 1/5 cm is 8/15 cm³.

The volume of a rectangular prism can be found using the formula: Volume = length x width x height. In this case, the length, width, and height of the prism are given as 10/3 cm, 4/5 cm, and 1/5 cm respectively. Replace these dimensions in the formula:

Volume = (10/3 cm) x (4/5 cm) x (1/5 cm) = 8/15 cm³.

Therefore, the volume of the rectangular prism is 8/15 cm³.

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Answer:

The calculated z- value = 1.479 <  1.645 at 0.10 or 90% level of significance.

The null hypothesis is accepted at 90% level of significance.

There is no significant difference in the proportion of E. Coli in organic vs. conventionally grown produce.

Step-by-step explanation:

Step:-(i)

Given first sample size n₁ = 200

The first sample proportion     [tex]p_{1} = \frac{5}{200} = 0.025[/tex]

Given first sample size n₂= 500

The second sample proportion     [tex]p_{2} = \frac{25}{500} = 0.05[/tex]

Step:-(ii)

Null hypothesis :H₀:There is no significant difference in the proportion of E. Coli in organic vs. conventionally grown produce

Alternative hypothesis:-H₁

There is  significant difference in the proportion of E. Coli in organic vs. conventionally grown produce

level of significance ∝=0.10

Step:-(iii)

          The test statistic    

                                        [tex]Z =\frac{p_{1} - p_{2} }{\sqrt{pq(\frac{1}{n_{1} }+\frac{1}{n_{2} } } }[/tex]

     where         p =   [tex]\frac{n_{1} p_{1} + n_{2}p_{2} }{n_{1}+n_{2} }= \frac{200X0.025+500X0.05 }{500+200}[/tex]

                     p = 0.0428

                     q = 1-p =1-0.0428 = 0.9572

                                      [tex]Z =\frac{0.025- 0.05}{\sqrt{0.0428X0.9571(\frac{1}{200 }+\frac{1}{500 } } }[/tex]

                                     Z = -1.479

                                    |z| = |-1.479|

                                    z = 1.479

The tabulated value z= 1.645 at 0.10 or 90% level of significance.

The calculated z- value = 1.479 <  1.645 at 0.10 or 90% level of significance.

The null hypothesis is accepted at 90% level of significance.

Conclusion:-

There is no significant difference in the proportion of E. Coli in organic vs. conventionally grown produce

Answer:

(ab - 10)(ab + 10)

Step-by-step explanation:

[tex] {a}^{2} {b}^{2} - 100 \\ = (ab) ^{2} - (10)^{2} \\ = (ab - 10)(ab + 10) \\ [/tex]