5(x+2)=⅗ (5+10x) how do i solve

A recent study of two vendors of desktop personal computers reported that out of 836 units sold by Brand A, 111 required repair,

iris [78.8K]

Answer:

Step-by-step explanation:

Hello!

The study variables are:

$X_A$ : The number of Brand A units sold that required repair.

$n_A= 836$

$x_A= 111$

$X_B$ : THe number of Brand B units sold that required repair.

$n_B= 739$

$x_B= 111$

1. Calculate the difference in the sample proportion for the two brands of computers, p^BrandA−p^BrandB =?.

The sample proportion of each sample is equal to the number of "success" observed xi divided by the sample size n:

^p $_A$ = $\frac{x_A}{n_A}= \frac{111}{836}= 0.1328$

^p $_B$ = $\frac{x_B}{n_B}= \frac{111}{739} =0.1502$

^p $_A$ - ^p $_B$ = 0.1328 - 0.1502= -0.0174

Note: proportions take numbers from 0 to 1, meaning they are always positive. But this time what you have to calculate is a difference between the two proportions so it is absolutely correct to reach a negative number it just means that one sample proportion is greater than the other.

2. What are the correct hypotheses for conducting a hypothesis test to determine whether the proportion of computers needing repairs is different for the two brands?

A. H0:pA−pB=0 , HA:pA−pB<0

B. H0:pA−pB=0 , HA:pA−pB>0

C. H0:pA−pB=0 , HA:pA−pB≠0

If you want to test whether the proportion of computers of both brands is different, you have to do a two-tailed test, the correct option is C.

3. Calculate the pooled estimate of the sample proportion, ^p= ?

To calculate the pooled sample proportion you have to use the following formula:

^p= $\frac{x_A+x_B}{n_A+n_B}= \frac{111+111}{836+739}= 0.14095 = 0.1410$

4. Is the success-failure condition met for this scenario?

A. Yes

B. No

The conditions that have to be met are:

$n_A\geq 30$ ⇒ Met

$n_A*p_A\geq 5$ ⇒ 836 * 0.1328= 111.4192; Met

$n_A*(1-p_A)\geq 5$ ⇒ 836 * (1 - 0.1328)= 727.5808; Met

$n_B\geq 30$ ⇒ Met

$n_B*p_B\geq 5$ ⇒ 739 * 0.1502= 110.9978; Met

$n_B*(1-p_B)\geq 5$ ⇒ 739 * (1-0.1502)= 628.0022; Met

All conditions are met.

5. Calculate the test statistic for this hypothesis test. ? =

$Z_{H_0}= \frac{(p'_A-p'_B)-(p_A-p_B)}{\sqrt{p'(1-p')[\frac{1}{n_A} +\frac{1}{n_B} ]} } = \frac{-0.0174-0}{\sqrt{0.1410*0.859*[\frac{1}{836} +\frac{1}{739} ]} }= -0.9902$

6. Calculate the p-value for this hypothesis test, p-value = .

This hypothesis test is two-tailed and so is the p-value, since it has two tails you have to calculate it as:

P(Z≤-0.9902) + P(Z≥0.9902)= P(Z≤-0.9902) + ( 1 - P(Z≤0.9902))= 0.161 + (1 - 0.839) = 0.322

7. What is your conclusion using α = 0.05?

A. Do not reject H0

B. Reject H0

The decision rule using th ep-value is:

If p-value > α, the decision is to not reject the null hypothesis.

If p-value ≤ α, the decision is to reject the null hypothesis.

The p-value= 0.322 is greater than α = 0.05, so the decision is to not reject the null hypothesis.

8. Compute a 95 % confidence interval for the difference p^BrandA−p^BrandB = ( , )

The formula to calculate the Confidence interval is a little different, because instead of the pooled sample proportion you have to use the sample proportion of each sample to calculate the standard deviation of the distribution:

( $p'_A-p'_B$ ) ± $Z_{1-\alpha /2}$ * $\sqrt{\frac{p'_A(1-p'_A)}{n_A} +\frac{p'_B(1-p'_B)}{n_B} }$

-0.0174 ± 1.965 * $\sqrt{\frac{0.1328*0.8672}{836} +\frac{0.1502*0.8498}{739} }$

[-0.0520; 0.0172]

I hope it helps!

3 0

3 years ago

A clinical psychologist wants to test whether experiencing childhood trauma reduces one's self-efficacy in adulthood. He randoml

lidiya [134]

Answer:

No, there is not any sufficient evidence to conclude that individuals who have experienced childhood trauma have lower self-efficacy in adulthood.

Step-by-step explanation:

We are given that a clinical psychologist wants to test whether experiencing childhood trauma reduces one's self-efficacy in adulthood.

He randomly selects 22 adults who have experienced childhood trauma and finds that their mean self-efficacy score equals 118.1.

Self-efficacy scores in the general population of adults are distributed normally with a mean equal to 118.5 and a standard deviation equal to 18.8 .

Let $\mu$ = mean self-efficacy score.

So, Null Hypothesis, $H_0$ : $\mu \geq$ 118.5 {means that the individuals who have experienced childhood trauma have higher or same self-efficacy in adulthood}

Alternate Hypothesis, $H_A$ : $\mu$ < 118.5 {means that the individuals who have experienced childhood trauma have lower self-efficacy in adulthood}

The test statistics that would be used here One-sample z test statistics as we know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean self-efficacy score = 118.1

$\sigma$ = population standard deviation = 18.8

n = sample of adults who have experienced childhood trauma = 22

So, test statistics = $\frac{118.1-118.5}{\frac{18.8}{\sqrt{22} } }$

= -0.0998

The value of z test statistics is -0.0998.

Since, in the question we are not given with the level of significance so we assume it to be 5%. Now, at 5% significance level the z table gives critical value of -1.645 for left-tailed test.

Since our test statistic is more than the critical value of z as -0.0998 > -1.645, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the individuals who have experienced childhood trauma have higher or same self-efficacy in adulthood.

4 0

4 years ago

Find the derivative of x^2+6x+9 at the point (2,6).

Pachacha [2.7K]

The derivative of the function is 2x+6.
At the particular point (2,6), plug in the x coordinate for x.
2(2)+6 = 10

Hope this helps :)

4 0

3 years ago

A craftsman builds two kinds of birdhouses, one for wrens and one for bluebirds. Each wren birdhouse takes 3 hours of labor and

Ksenya-84 [330]

Answer:

Obj Function, Max P=8w+16b

Subject to the Constraints

3w+2l≤80

4w+10b≤100

w≥6

b>0

Step-by-step explanation:

Let the number of Wren's birdhouse built=w

Let the number of bluebirds birdhouse built=b

Constraint:Labour

Each wren birdhouse takes 3 hours of labor.

Each bluebird house requires 2 hours of labor.

The craftsman has available 80 hours of labor

Formulated as an inequality:

3w+2l≤80

Constraint: Lumber

Each wren birdhouse requires 4 units of lumber.

Each bluebird birdhouse 10 units of lumber.

The craftsman has available 100 units of lumber

Formulated as an inequality:

4w+10b≤100

The craftsman wants to build at least 6 wren houses.

w≥6

Since he builds the two kinds of house, b>0 and w>0.

Objective Function

Wren houses profit $8 each and bluebird houses profit $16 each.

The Objective of the function is to make profit, so we maximize.

Obj Function, Max P=8w+16b

The linear programming problem is therefore stated below:

Obj Function, Max P=8w+16b

Subject to the Constraints

3w+2l≤80

4w+10b≤100

w≥6

b>0

4 0

3 years ago

The measure of angle θ is 7pi/4 . The measure of its reference angle is__ °, and tan θ is__

irga5000 [103]

To get the reference angle, convert the angle measurement to degrees first since it is in radians.

The conversion factor is pi = 180°

So, $\frac{7pi}{4} x \frac{180}{pi}$
$\frac{`1260}{4} = 315$

Since 315° is located in quadrant IV, to get the reference angle, subtract 315° from 360°. Hence, the reference angle is 45°.

Based on the basics of trigonometric functions, Tan (45) = 1.

5 0

4 years ago

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