A bread recipe calls for 1 teaspoon of yeast for every 2
cups of flour.
Write an equation that represents the number of cups of
flour, c, for every teaspoon of yeast, t.

The volume of the sphere Y is 27 times the volume of the sphere X

Step-by-step explanation:

Sphere X:

Radius of sphere X = 3 inches

Volume = (4/3) π(27)

Sphere Y:

Radius of sphere Y = 9 inches

Volume = (4/3) π(729)

Volume of sphere Y /volume of sphere X = (4/3) π(729) /(4/3) π(27)

= (729) /(27)

= 27

The volume of the sphere Y is 27 times the volume of the sphere X

Answer:

$1.98/ lbs

Step-by-step explanation:

To find the unit cost, take the total cost and divide the the number of units purchased

$9.90/5 lbs

$1.98/ lbs

Answer:

The base area is the total area of D

The lateral area is the total area of A, B, and C

The surface area is the total area of A, B, C, and D.

Answer:

(d) the base area is the total area

Explain: because i got it right on the test

The best prediction for selecting a blue marble 60 times, with replacement, from a bag of 4 blue marbles and 3 white marbles is approximately 34.29 times.

To predict the number of times you will select a blue or purple marble, we need to know the total number of marbles, the number of blue marbles, the number of purple marbles, and the number of trials (which is 60 in this case). Since no information about purple marbles is provided, let's assume you meant blue marbles only for this explanation. In the scenario presented in the question, the probability of drawing a blue marble is 4/7 each time since there are 4 blue marbles and 3 white marbles, totaling 7 marbles.

Therefore, the best prediction for the number of times a blue marble will be selected is approximately 34.29 times (which you might round to 34 times).

Answer:

B

Step-by-step explanation:

this is true because if you rotate 180 degrees on a graph both the x and y will always become their opposite.

Answer:

Part A: The null hypothesis failed to be rejected.

Part B: The null hypothesis failed to be rejected.

Step-by-step explanation:

We have an hypothesis test with null and alternative hypothesis H0: p = 0.5 versus Ha: p > 0.5, which has the test statistic z=1.15.

Part A: If the significance level is 0.05, the conclusion depends on the P-value.

If the P-value is below 5%, the null hypothesis is rejected.

The P-value for this right-tailed tes and z=1.15 is:

[tex]P-value=P(z>1.15)=0.125[/tex]

The P-value is bigger than the significance level, so the effect is not significant and the null hypothesis is failed to be rejected.

Part B: In this case, the significance level is 0.01 and, as the alternative hypothesis is defined with an unequal sign, the test is two-tailed.

This changes the way we calculate the P-value, as we need to compute the two tails.

The P-value is:

[tex]P-value=2P(z>1.15)=0.25[/tex]

The P-value is bigger than the significance level, so the effect is not significant and the null hypothesis is failed to be rejected.

Null hypothesis of a sample suggest that there is no statistical relationship exists in a set of provided single observed variable. The null hypothesis is failed to be rejected for both part A and part B of the question.

Using the hypothesis test we have,

[tex]H_o; p=0.5[/tex]

[tex]H_ap>0.5[/tex]

[tex]z=1.15[/tex]

Null hypothesis of a sample suggest that there is no statistical relationship exists in a set of provided single observed variable.

Part A: The conclusion to draw at the 5% significance level-

In the right tailed test or upper test the p value for the z score can be given form z table which is,

p-value[tex]=0.125[/tex]

Here the p value 0.125 is bigger than the significance level 0.05. Thus the effect is not significant and the null hypothesis is failed to be rejected.

Part B: If the alternative hypothesis is Ha: p ≠ 0.5, the conclusion to draw at the 5% significance level-

Here for the unequal sign 2-test is performed. The p value for the 2-test is,

p- value[tex]=0.25[/tex]

Here the p value 0.25 is bigger than the significance level 0.05. Thus the effect is not significant and the null hypothesis is failed to be rejected.

Hence, the null hypothesis is failed to be rejected for both part A and part B of the question.

For more about the null hypothesis, follow the link below-

https://brainly.com/question/4454077

Answer:36.36feet (that rounded but if you want you can round it to 36.4

Step-by-step explanation:

Final answer:

The problem involves using the Pythagorean theorem to find the length of a kite string given the horizontal and vertical distance, which would be the square root of the sum of the squares of 19 feet and 31 feet.

Explanation:

The student's question concerns the length of the string used to fly a kite when given the horizontal distance and the vertical distance from the ground to the kite. To solve this, we apply the Pythagorean theorem, which states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. In this problem, the kite string functions as the hypotenuse, the horizontal distance to the kite is one leg, and the vertical distance from the ground to the kite is the other leg.

For Mike's kite:

According to the Pythagorean theorem:

[tex]\( c^2 = a^2 + b^2 \\( c^2 = 19^2 + 31^2 \\( c^2 = 361 + 961 \\( c^2 = 1322 \\( c = \\( c \\) = \\( c \\) = \\) is the length of the kite string[/tex].

Therefore, Mike's kite string is approximately \\( c \\) long.

Answer:

The number of standard deviations from $1,158 to $1,360 is 1.68.

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 = 1158, \sigma = 120[/tex]

The number of standard deviations from $1,158 to $1,360 is:

This is Z when X = 1360. So

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

[tex]Z = \frac{1360 - 1158}{120}[/tex]

[tex]Z = 1.68[/tex]

The number of standard deviations from $1,158 to $1,360 is 1.68.

Answer:

25/36

Step-by-step explanation:

P(1) = 1/6

P'(1) = 5/6

5/6 * 5/6 = 25/36

Answer:

x = 7

Step-by-step explanation:

[tex]m\overset{\frown} {PQ} = m\overset{\frown} {MN}... (given) \\

\therefore Chord \:PQ \cong Chord\: MN\\

\therefore \: 4x + 10 = 38 \\ \therefore \: 4x= 38 - 10\\ \therefore \: 4x= 28\\ \therefore \: x= \frac{28}{4} \\ \huge \red{ \boxed{\therefore \: x= 7}}[/tex]

Answer:

The confidence interval for the proportion of students supporting the fee increase

( 0.77024, 0.81776)

Step-by-step explanation:

Explanation:

Given data a survey of an urban university (population of 25,450) showed that 883 of 1,112 students sampled supported a fee increase to fund improvements to the student recreation center.

Given sample size 'n' = 1112

Sample proportion 'p' = [tex]\frac{883}{1112} = 0.7940[/tex]

                           q = 1 - p = 1- 0.7940 = 0.206

The 95% level of confidence intervals

The confidence interval for the proportion of students supporting the fee increase

[tex](p-z_{\alpha } \sqrt{\frac{pq}{n} } ,p + z_{\alpha } \sqrt{\frac{pq}{n} } )[/tex]

The Z-score at 95% level of significance =1.96

[tex](0.7940-1.96\sqrt{\frac{0.7940 X 0.206}{1112} } ,0.7940 + 1.96 \sqrt{\frac{0.7940 X 0.206}{1112} } )[/tex]

(0.7940-0.02376 , 0.7940+0.02376)

( 0.77024, 0.81776)

Conclusion:-

The confidence interval for the proportion of students supporting the fee increase

( 0.77024, 0.81776)

Answer: I think the answer is B. I don't fully understand the question but it seems like that would be the answer. You might double check though.

Step-by-step explanation:

Answer:

A. The statement that the random variable X is normally distributed with parameters is often abbreviated.

Step-by-step explanation:

The X is normally distributed with parameters u and  is often distributed as X ≈N (u 2 ). The correct answer is a. The statement that the random variable is normally distributed.

Answer: 283cm

Step-by-step explanation:

Answer:

Expected value of the total volume shipped = 97,650 ft³

Variance of the total volume shipped = 15,183,265

Standard deviation = 3896.6 ft³

Step-by-step explanation:

The mean of number of type 1, 2 and 3 containers in a week

μ₁ = 230, μ₂ = 240, μ₃ = 120

The standard deviations for the number of type 1, 2 and 3 containers in a week

σ₁ = 11, σ₂ = 12, σ₃ = 7

When independent distributions are combined, the combined mean and combined variance are given through the relation

Combined mean = Σ λᵢμᵢ

(summing all of the distributions in the manner that they are combined)

Combined variance = Σ λᵢ²σᵢ²

(summing all of the distributions in the manner that they are combined)

Volume of each container type

λ₁ = 27 ft³

λ₂ = 125 ft³

λ₃ = 512 ft³

Distribution of total volume shipped

= 27X₁ + 125X₂ + 512X₃

Expected value = Combined Mean = 27μ₁ + 125μ₂ + 512μ₃

= (27×230) + (125×240) + (512×120) = 590

Combined Variance = 27²σ₁² + 125²σ₂² + 512²σ₃²

= (27² × 11²) + (125² × 12²) + (512² × 7²)

= 88,209 + 2,250,000 + 12,845,056

= 15,183,265

Standard deviation = √(15,183,265) = 3896.6 ft³

Hope this Helps!!!

Answer:

111

Step-by-step explanation:

-The occupancy rate index is calculated using the following formula:

[tex]Index=\frac{Hotel \ OCC}{Competitive \ OCC}\times 100[/tex]

-The Hotel OCC is the property's occupancy rate given as 85%

-The Competitive OCC, is the competitive set's occupancy rate calculated as;

[tex]Occupancy \ Rate=\frac{Room \ Sold}{Rooms \ Available}\\\\\\=\frac{115000}{150000}\\\\=76.67\%[/tex]

-We then substitute to solve for the occupancy rate index:

[tex]Index=\frac{Hotel \ OCC}{Competitive \ OCC}\times 100\\\\=\frac{0.85}{0.7767}\times 100\\\\=110.86\\\approx 111[/tex]

Hence, the occupancy rate index is 111

Answer:

[tex]P(X\geq 5)=1-P(X<5)=1-P(X\leq 4)=1-[P(X=0)+P(X=1)+P(X=2) +P(X=3) +P(X=4)][/tex]

Using the pmf we can find the individual probabilities like this:

[tex]P(X=0)=\frac{e^{-6.4} 6.4^0}{0!}=0.001662[/tex]

[tex]P(X=1)=\frac{e^{-6.4} 6.4^1}{1!}=0.010634 [/tex]

[tex]P(X=2)=\frac{e^{-6.4} 6.4^2}{2!}=0.034029 [/tex]

[tex]P(X=3)=\frac{e^{-6.4} 6.4^3}{3!}=0.072595 [/tex]

[tex]P(X=4)=\frac{e^{-6.4} 6.4^4}{4!}=0.116151 [/tex]

And replacing we got:

[tex] P(X \geq 5) =0.76493[/tex]

Step-by-step explanation:

Previous concepts

Let X the random variable that represent the number of people that will become sereiosly ill in two years. We know that [tex]X \sim Poisson(\lambda)[/tex]

The probability mass function for the random variable is given by:

[tex]f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...[/tex]

And f(x)=0 for other case.

For this case the value for [tex]\lambda[/tex] would be:

[tex]\lambda = 3.2 \frac{ills}{year} *2 years = 6.4[/tex]

For this distribution the expected value is the same parameter [tex]\lambda[/tex]

[tex]E(X)=\mu =\lambda[/tex]

On this case we are interested on the probability of having at least two chocolate chips, and using the complement rule we have this:

[tex]P(X\geq 5)=1-P(X<5)=1-P(X\leq 4)=1-[P(X=0)+P(X=1)+P(X=2) +P(X=3) +P(X=4)][/tex]

Using the pmf we can find the individual probabilities like this:

[tex]P(X=0)=\frac{e^{-6.4} 6.4^0}{0!}=0.001662[/tex]

[tex]P(X=1)=\frac{e^{-6.4} 6.4^1}{1!}=0.010634 [/tex]

[tex]P(X=2)=\frac{e^{-6.4} 6.4^2}{2!}=0.034029 [/tex]

[tex]P(X=3)=\frac{e^{-6.4} 6.4^3}{3!}=0.072595 [/tex]

[tex]P(X=4)=\frac{e^{-6.4} 6.4^4}{4!}=0.116151 [/tex]

And replacing we got:

[tex] P(X \geq 5) =0.76493[/tex]

Answer:

The probability that at least 5 people will become seriously ill in two years is 0.7649.

Step-by-step explanation:

The question mentions that this problem can be modeled using the Poisson distribution so, we will use the formula:

P(X=x) = [(e^-λt)*(λt^x)]/x!

where λ = average number of occurrences per year

           t = no. of years

           x = number of people

We need to determine P(X≥5) so first we will calculate the probabilities at X=0,1,2,3,4 and subtract them from the total probability i.e. 1 to find P(X≥5).

We have λ = 3.2, t=2 years hence λt = (3.2)(2) = 6.4. So,

P(X=0) = [(e^(-6.4)*(6.4^0)]/0! = 0.00166

P(X=1) = [(e^(-6.4)*(6.4^1)]/1! = 0.01063

P(X=2) = [(e^(-6.4)*(6.4^2)]/2! = 0.03403

P(X=3) = [(e^(-6.4)*(6.4^3)]/3! = 0.07259

P(X=4) = [(e^(-6.4)*(6.4^4)]/4! = 0.011615

P(X≥5) = 1 - P(X<5)

           = 1 - [P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4)]

           = 1 - (0.00166 + 0.01063 + 0.03403 + 0.07259 + 0.011615)

           = 1 - 0.23506

P(X≥5) = 0.7649

The probability that at least 5 people will become seriously ill in two years is 0.7649.

Answer:

0.6915

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 = 88, \sigma = 4[/ex]

What is the probability that a single car of this model emits more than 86 mg/mi of NOX + NMOG?

This is 1 subtracted by the pvalue of Z when X = 86. So

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

[tex]Z = \frac{86 - 88}{4}[/tex]

[tex]Z = -0.5[/tex]

[tex]Z = -0.5[/tex] has a pvalue of 0.3085

1 - 0.3085 = 0.6915

The answer is 0.6915

Answer:

4=c

Step-by-step explanation:

7c + 3 - 7c = 3c - 9​

Combine like terms

7c-7c+3 = 3c-9

3 = 3c-9

Add 9 to each side

3+9 = 3c-9+9

12 = 3c

Divide each side by 3

12/3 = 3c/3

4 =c

Answe

c=−4/3

Step-by-step explanation:

Look at the attached picture⤴

Hope it will help uh...:)

Answer:

Mean: 57.667

Median: 59

Mode: 83

Range: 71

Explanation

Mean = sum of values / number of values

Median = middle number

Mode = number that occurs most often

Range = largest number minus smallest number

Answer:  (a). Passwords = 48620250

               (b). Passwords = 3111696000

Step-by-step explanation:

The question reads that 'A certain organization recommends the use of passwords with the following format':

Consonant, Vowel, Consonant, Consonant, Vowel, Consonant, Number, Number (for example, pinray45​).

The number of consonants is 21, the number of vowels is 5 and the numbers are 10 (0 to 9).

Since, there are 21 consonants and 5 vowels and repetition is allowed.

Use the general multiplication rule for each characters in the sequence.

Multiply by 21 consonants, 5 for vowels and 10 for numbers.

Assuming that the passwords are not case sensitive.

i.e. upper case & lower case letters are same.

Thus, the number of possible passwords is:

Passwords = Consonant * Vowel * Consonant * Consonant * Vowel * Consonant * Number * Number

Passwords = 5 x 21 x 21 x 21 x 5 x21 x 10 = 48620250

(b). Assuming that the passwords are case sensitive

Passwords = 10 x 42 x 42 x 42 x 10 x 42 x 10 = 42 x 420³ = 3111696000

cheers i hope this helps!!!!

The total number of possible passwords is 95,887,125,000.

To calculate the number of possible passwords, we need to consider the number of choices for each position in the password. In the given format, we have 5 positions for consonants, 2 positions for vowels, and 1 position for a number. Let's break it down:

For the 1st consonant position, we have 21 choices (since there are 21 consonants).

For the 1st vowel position, we have 5 choices (since there are 5 vowels).

For the 2nd consonant position, we have 21 choices.

For the 2nd vowel position, we have 5 choices.

For the 3rd consonant position, we have 21 choices.

For the number position, we have 10 choices (since there are 10 numbers).

For the 4th consonant position, we have 21 choices.

For the 3rd vowel position, we have 5 choices.

For the 5th consonant position, we have 21 choices.

Finally, for the last consonant and the last vowel positions, we have 21 and 5 choices respectively.

To find the total number of possible passwords, we multiply the number of choices for each position:

Total number of passwords = 21 * 5 * 21 * 5 * 21 * 10 * 21 * 5 * 21 * 5 = 95,887,125,000.

Answer:

Mathematical  200

reasonable 0,200

Step-by-step explanation:

i got it right on the test

Answer:

2.2

Step-by-step explanation:

It would take 15840 min to eat all of them that means 264 hours or 11 days

lol why you need this info?

Answer:

  16

Step-by-step explanation:

You want the mean of {20, 15, 19, 16, 10}.

The mean is the sum of the numbers, divided by the number of numbers.

  (20 +15 +19 +16 +10)/5 = 80/5 = 16

The mean is 16.

Final answer:

The mean of tickets sold by Helena for the Willowbrook School play is calculated by adding the total tickets sold (80) and dividing by the number of homerooms (5), resulting in a mean of 16 tickets.

Explanation:

To calculate the mean of tickets sold for the Willowbrook School play, we need to add up the number of tickets sold in each homeroom and then divide by the number of homerooms. The data given is:

Adding these together, we get 20 + 15 + 19 + 16 + 10 = 80 tickets sold in total. As there are 5 sets of data, we divide the total by 5 to find the mean.

Therefore, the mean number of tickets sold is 16 tickets.

Answer:

[tex]P(X<72)=P(\frac{X-\mu}{\sigma}<\frac{72-\mu}{\sigma})=P(Z<\frac{72-80}{5})=P(z<-1.6)[/tex]

And we can find this probability with the normal standard distirbution or excel and we got:

[tex]P(z<-1.6)=0.055[/tex]

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the lenghts of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(80,5)[/tex]  

Where [tex]\mu=80[/tex] and [tex]\sigma=5[/tex]

We are interested on this probability

[tex]P(X<72)[/tex]

And the best way to solve this problem is using the normal standard distribution and the z score given by:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

If we apply this formula to our probability we got this:

[tex]P(X<72)=P(\frac{X-\mu}{\sigma}<\frac{72-\mu}{\sigma})=P(Z<\frac{72-80}{5})=P(z<-1.6)[/tex]

And we can find this probability with the normal standard distirbution or excel and we got:

[tex]P(z<-1.6)=0.055[/tex]

Answer:

B

Step-by-step explanation:

Answer:

24

Step-by-step explanation:

Answer:

The answer is C. x=In(5/3)-2

Step-by-step explanation:

Answer:

C

Step-by-step explanation: