Solve for the variable. -2x = -15

Wyatt plays on the school baseball team. In the last 10 games, Wyatt was

Ksju [112]

Answer:

Step-by-step explanation:

21 hits / 45 at bats

7/15

5 0

3 years ago

After once again losing a football game to the college'ss arch rival, the alumni association conducted a survey to see if alumni

antoniya [11.8K]

Answer:

a) The 90% confidence interval would be given (0.561;0.719).

b) $p_v =P(z>2.8)=1-P(z$

c) Using the significance level assumed $\alpha=0.01$ we see that $p_v$ so we have enough evidence at this significance level to reject the null hypothesis. And on this case makes sense the claim that the proportion is higher than 0.5 or 50%.

Step-by-step explanation:

1) Data given and notation

n=100 represent the random sample taken

X=64 represent were in favor of firing the coach

$\hat p=\frac{64}{100}=0.64$ estimated proportion for were in favor of firing the coach

$p_o=0.5$ is the value that we want to test since the problem says majority

$\alpha$ represent the significance level (no given, but is assumed)

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

p= population proportion of Americans for were in favor of firing the coach

Part a

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 90% confidence interval the value of $\alpha=1-0.9=0.1$ and $\alpha/2=0.05$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.64$

And replacing into the confidence interval formula we got:

$0.64 - 1.64 \sqrt{\frac{0.64(1-0.64)}{100}}=0.561$

$0.64 + 1.64 \sqrt{\frac{0.64(1-0.64)}{100}}=0.719$

And the 90% confidence interval would be given (0.561;0.719).

Part b

We need to conduct a hypothesis in order to test the claim that the proportion exceeds 50%(Majority). :

Null Hypothesis: $p \leq 0.5$

Alternative Hypothesis: $p >0.5$

We assume that the proportion follows a normal distribution.

This is a one tail upper test for the proportion of union membership.

The One-Sample Proportion Test is "used to assess whether a population proportion $\hat p$ is significantly (different,higher or less) from a hypothesized value $p_o$ ".

Check for the assumptions that he sample must satisfy in order to apply the test

a)The random sample needs to be representative: On this case the problem no mention about it but we can assume it.

b) The sample needs to be large enough

$np_o =100*0.64=64>10$

$n(1-p_o)=100*(1-0.64)=36>10$

Calculate the statistic

The statistic is calculated with the following formula:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o(1-p_o)}{n}}}$

On this case the value of $p_o=0.5$ is the value that we are testing and n = 100.

$z=\frac{0.64 -0.5}{\sqrt{\frac{0.5(1-0.5)}{100}}}=2.8$

The p value for the test would be:

$p_v =P(z>2.8)=1-P(z$

Part c

Using the significance level assumed $\alpha=0.01$ we see that $p_v$ so we have enough evidence at this significance level to reject the null hypothesis. And on this case makes sense the claim that the proportion is higher than 0.5 or 50%.

6 0

3 years ago

13+7 POINTS!!! PLZ HELP WILL GIVE BRAINLIEST!!!

adelina 88 [10]

If x=-5 (-5)2+2=27
likewise put -4 -3 -2 -1 0 1 2 3
and in another equation also.

6 0

3 years ago

How do you do this u don't understand

Elden [556K]

You have to make the denominator the same: 40
35/40 + 32/40 = 67/40

7 0

3 years ago

PLEASE HELP ME!!

Ludmilka [50]

Answer:

<h2>The population will reach 1200 after about 2.8 years</h2>

Step-by-step explanation:

The question is incomplete. Here is the complete question.

The population of a certain species of bird in a region after t years can be modeled by the function P(t) = 1620/ 1+1.15e-0.42t , where t ≥ 0. When will the population reach 1,200?

According to question we are to calculate the time t that the population P(t) will reach 1200.To do this we will substitute P(t) = 1,200 into the equation and calculate for the time 't'.

Given;

$P(t) = \frac{1620}{1+1.15e^{-0.42t} } \\\\at \ P(t)= 1200;\\\\1200 = \frac{1620}{1+1.15e^{-0.42t} }\\\\cross\ multiplying\\\\1+1.15e^{-0.42t} = \frac{1620}{1200} \\\\1+1.15e^{-0.42t} = 1.35\\\\1.15e^{-0.42t} = 1.35-1\\\\e^{-0.42t} = \frac{0.35}{1.15}\\ \\e^{-0.42t} = 0.3043\\\\Taking \ ln\ of\ both\ sides\\\\lne^{-0.42t} = ln0.3043\\\\-0.42t = -1.1897\\\\t = \frac{-1.1897}{-0.42} \\\\t = 2.8 years\\\\$

The population will reach 1200 after about 2.8 years

6 0

3 years ago