NEED HELP ASAP PLZZZZ

Write an equation for this tape diagram 6th grade math help pls thanks if you do

Tatiana [17]

Answer: x(15x)

Step-by-step explanation:

4 0

2 years ago

Harris Interactive® conducted a poll of American adults in August of 2011 to study the use of online medical information. Of the

kirill115 [55]

Answer:

$0.60 - 1.96\sqrt{\frac{0.60(1-0.60)}{1019}}=0.570$

$0.60 + 1.96\sqrt{\frac{0.60(1-0.60)}{1019}}=0.630$

The 95% confidence interval for the true proportion would be given by (0.570;0.630) .

And if we convert this into % we got (57.0%, 63.0%)

Step-by-step explanation:

The information given we have the following info given:

$n = 1019$ represent the sampel size

$\hat p=0.6$ represent the sample proportion of interest

The confidence level is 95%, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

The confidence interval for the mean is given by the following formula:

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

Replacing the info given we got:

$0.60 - 1.96\sqrt{\frac{0.60(1-0.60)}{1019}}=0.570$

$0.60 + 1.96\sqrt{\frac{0.60(1-0.60)}{1019}}=0.630$

The 95% confidence interval for the true proportion would be given by (0.570;0.630) .

And if we convert this into % we got (57.0%, 63.0%)

7 0

2 years ago

Please help me? I'm almost certain it is only option A but I want a second opinion.

jek_recluse [69]

Option c compounded daily means more interest

8 0

3 years ago

Read 2 more answers

How does math anxiety affect working memory, and why do such changes in

Alika [10]

Anxiety can cause you to overthink what you already know which interns makes you think you don’t know it. Math is just formulas and arthritic and if anxiety has you doubting that you know how to solve a problem then you won’t remember how to solve it

6 0

2 years ago

A digital camcorder repair service has set a goal not to exceed an average of 5 working days from the time the unit is brought i

garri49 [273]

Answer:

The degrees of freedom first given by:

$df=n-1=12-1=11$

Then we can find the critical value taking in count the degrees of freedom and the alternative hypothesis and then we need to find a critical value who accumulates 0.05 of the area in the right tail and we got:

$t_{\alpha}= 1.796$

And for this case the rejection region would be:

b) Reject H0 if tcalc >1.7960

Step-by-step explanation:

Information given

5, 7, 4, 6, 7, 5, 5, 6, 4, 4, 7, 5.

System of hypothesis

We want to test if the true mean is higher than 5, the system of hypothesis are :

Null hypothesis: $\mu \leq 5$

Alternative hypothesis: $\mu > 5$

The statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

The degrees of freedom first given by:

$df=n-1=12-1=11$

Then we can find the critical value taking in count the degrees of freedom and the alternative hypothesis and then we need to find a critical value who accumulates 0.05 of the area in the right tail and we got:

$t_{\alpha}= 1.796$

And for this case the rejection region would be:

b) Reject H0 if tcalc >1.7960

6 0

3 years ago