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2 years ago

When Quinn got home, he turned the air conditioner on.

Mathematics

1 answer:

aleksley [76]2 years ago

3 0

Answer:

It’s 42

Step-by-step explanation: when Quinn return home, it was 0 minutes he turned on the air conditioner. So temperature at that time is equal to value of Q when t=0

Q=42-0.7t

=42-0.7(0)

=42

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A blimp is 1500 meters high in the air and measures the angles of depression to two stadiums to the west of the blimp. If those

grigory [225]

Let stadium 1 be the one on the left and stadium 2 the one on the right.

Angle above stadium 1 is 72.9° and the angle above stadium 2 is 34.1° using the angle property of alternate angles(because both the ground and the dotted line are parallel).

For the next part we need to use the trigonometric function of tangent.

As tan x = opposite / adjacent,
Tan 72.9°=1500/ adjacent ( the ground from O to stadium 1)

Therefore the adjacent is 1500/tan 72.9°= 461.46 m( to 5 s.f.)

Same for the next angle,
Tan 34.1°=1500/ adjacent ( the ground from O to stadium 2)

Therefore, the adjacent is 1500/tan 34.1° = 2215.49 m (to 5 s.f.)

Thus, the distance between both stadiums is 2215.49-461.46= 1754.03 m

Correcting the answer to whole number gives you 1754 m which is the option C.

4 0

2 years ago

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

The average credit card debt for a recent year was $8,776. Five years earlier the average credit card debt was $8,189. Assume sa

Anestetic [448]

Answer:

We reject H₀. We support that the new average credit card debt is bigger than the previous average

Step-by-step explanation:

Five years earlier

μ = 8189

σ = 690

Sample size n = 32

Recent year debt

x = 8776

Sample size n = 32

a) Hypothesis Test:

Null Hypothesis H₀ x = μ = 8189

Alternative Hypothesis Hₐ x > μ

b) z(c) Alternative Hypothesis establishes that the test is a one tail-test to the right.

z(c) for significance level α = 0.05 is from z-table z(c) = 1,64

c) z(s) = ( x - μ ) / σ /√n

z(s) = ( 8776 - 8189 ) / 690 /√32

z(s) = 587 *5,66/ 690

z(s) = 4,81

d) Comparing z(c) and z(s)

z(s) > z(c) Then z(c) is in the rejection region and we reject H₀

e) We have evidence that at 95 % of confidence the new value for the debt in credit card is now bigger than the average

7 0

1 year ago

The traffic engineer for a large city is conducting a study on traffic flow at a certain intersection near the city administrati

barxatty [35]

Answer:

e) The number of minutes for a car to get from the intersection to the administration building

Step-by-step explanation:

Continuous Data:

Data that can take any value (an infinite number of values) within a certain range.

For example, the statistics of a group of people form continuous data, but the number of people in that group form discrete data.

Inglés

In this case, counting items such as cars, bicycles, people, are considered discrete data. They exclusively take integer values.

But time data can take continuous values.

8 0

2 years ago

The formula to determine energy is uppercase E = one-half m v squared. What is the formula solved for v?

alexandr402 [8]

Answer:

Step-by-step explanation:

$E=\frac{1}{2}mv^2$

All the variables on the right are being multiplied together then the whole mess is being divided by 2. Let's get rid of the 2 first. The undoing of division is multiplication, so we will begin by multiplying both sides by 2 to get

$2E=mv^2$

Next we will move the m. The undoing of multiplication is division. So we divide both sides by m to get

$\frac{2E}{m}=v^2$

The undoing of a square is to take the square root, so we will do that to both sides giving us, finally

$\sqrt{\frac{2E}{m} }=v$

6 0

2 years ago

Read 2 more answers