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valentina_108 [34]

2 years ago

4

Cassie bought 2 apples for $4. Olivia bought 5 apples for $15. How much more money did Oliva spend per apple? Use the dollar sig

n!

1 answer:

gladu [14]2 years ago

8 0

Answer:

$1

Step-by-step explanation:

To find out how much more money Olivia spent per apple, we need to find out how much each spent on one apple.

Cassie:

2 apples --- $4

1 apple --- $4 ÷ 2 = $2

Olivia:

5 apples --- $15

1 apple --- $15 ÷ 5 = $3

Difference per apple = $3 - $2

= $1

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Employment data at a large company reveal that 74% of the workers are married, 42% are college graduates, and that 56% are marri

ValentinkaMS [17]

Answer:

(E) None of these above are true.

Step-by-step explanation:

Married = 74% or 0.74

College graduates = 42% or 0.42

pr(married | college graduates) = 0.56

(A) These events are pairwise disjoint. This is false. Pairwise disjoint are also known as mutually exclusive events. Here we can see that both events are occurring at same time.

(B) These events are independent events. This is also false.

(C) These events are both independent and pairwise disjoint. False

(D) A worker is either married or a college graduate always. False

Here Probability(A or B) shall be 1

= Pr(A) + Pr(B) - Pr( A and B) = 0.74 + 0.42 - 0.56 * 0.42 = 0.9248

This is not equal to 1.

(E) None of these above are true. This is true.

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2 years ago

Jim is building a rectangular deck and wants the length to be 1 ft greater than the width. what will be the dimensions of the de

Travka [436]

Let x = width
x+1 is then the length

2x+2(x+1)=66
2x+2x+2=66
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x=16
deck will be 16x17, nice for a BBQ. :)

3 0

2 years ago

The equation y=−0.0088x2+0.79x+15 models the speed x (in miles per hour) and average gas mileage y (in miles per gallon) for a v

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Answer: 30.72 miles per gallon

To find <span>best approximate for the average gas mileage at a speed of 60 miles per hour you need to replace the variable x with 60. The calculation would be:

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2 years ago

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Katie has 4 grape candies and 3 cherry candies. The ratio of the mumber of grape candies to the total number of candies is what?

dimaraw [331]

Answer 4:7

Step-by-step explanation:

6 0

1 year ago

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The heat evolved in calories per gram of a cement mixture is approximately normally distributed. The mean is thought to be 100,

Gre4nikov [31]

Answer:

A.the type 1 error probability is $\mathbf{\alpha = 0.0244 }$

B. β = 0.0122

C. β = 0.0000

Step-by-step explanation:

Given that:

Mean = 100

standard deviation = 2

sample size = 9

The null and the alternative hypothesis can be computed as follows:

$\mathtt{H_o: \mu = 100}$

$\mathtt{H_1: \mu \neq 100}$

A. If the acceptance region is defined as $98.5 < \overline x > 101.5$ , find the type I error probability $\alpha$ .

Assuming the critical region lies within $\overline x < 98.5$ or $\overline x > 101.5$ , for a type 1 error to take place, then the sample average x will be within the critical region when the true mean heat evolved is $\mu = 100$

∴

$\mathtt{\alpha = P( type \ 1 \ error ) = P( reject \ H_o)}$

$\mathtt{\alpha = P( \overline x < 98.5 ) + P( \overline x > 101.5 )}$

when $\mu = 100$

$\mathtt{\alpha = P \begin {pmatrix} \dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}} < \dfrac{\overline 98.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} + \begin {pmatrix}P(\dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}} > \dfrac{101.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} }$

$\mathtt{\alpha = P ( Z < \dfrac{-1.5}{\dfrac{2}{3}} ) + P(Z > \dfrac{1.5}{\dfrac{2}{3}}) }$

$\mathtt{\alpha = P ( Z 2.25) }$

$\mathtt{\alpha = P ( Z$

From the standard normal distribution tables

$\mathtt{\alpha = 0.0122+( 1- 0.9878) })$

$\mathtt{\alpha = 0.0122+( 0.0122) })$

$\mathbf{\alpha = 0.0244 }$

Thus, the type 1 error probability is $\mathbf{\alpha = 0.0244 }$

B. Find beta for the case where the true mean heat evolved is 103.

The probability of type II error is represented by β. Type II error implies that we fail to reject null hypothesis $\mathtt{H_o}$

Thus;

β = P( type II error) - P( fail to reject $\mathtt{H_o}$ )

∴

$\mathtt{\beta = P(98.5 \leq \overline x \leq 101.5) }$

Given that $\mu = 103$

$\mathtt{\beta = P( \dfrac{98.5 -103}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-103}{\dfrac{2}{\sqrt{9}}}) }$

$\mathtt{\beta = P( \dfrac{-4.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-1.5}{\dfrac{2}{3}}) }$

$\mathtt{\beta = P(-6.75 \leq Z \leq -2.25) }$

$\mathtt{\beta = P(z< -2.25) - P(z < -6.75 )}$

From standard normal distribution table

β = 0.0122 - 0.0000

β = 0.0122

C. Find beta for the case where the true mean heat evolved is 105. This value of beta is smaller than the one found in part (b) above. Why?

$\mathtt{\beta = P(98.5 \leq \overline x \leq 101.5) }$

Given that $\mu = 105$

$\mathtt{\beta = P( \dfrac{98.5 -105}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-105}{\dfrac{2}{\sqrt{9}}}) }$

$\mathtt{\beta = P( \dfrac{-6.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-3.5}{\dfrac{2}{3}}) }$

$\mathtt{\beta = P(-9.75 \leq Z \leq -5.25) }$

$\mathtt{\beta = P(z< -5.25) - P(z < -9.75 )}$

From standard normal distribution table

β = 0.0000 - 0.0000

β = 0.0000

The reason why the value of beta is smaller here is that since the difference between the value for the true mean and the hypothesized value increases, the probability of type II error decreases.

8 0

2 years ago