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3 years ago

15

Can I get help with 1 & 2??

1 answer:

Rashid [163]3 years ago

3 0

A. g(10)= -29
b.f(3)=16
c.h(-2)=-6
d.j(7)=23
e.h(a)=12/a
f. g(b+c)=-3b-3c+1 ...etc I'm bored

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A simple random sample of 20 days in which Parsnip ate seeds was selected, and the mean amount of time it took him to eat the se

bearhunter [10]

Answer:

$(14.5-13.9) -1.69 \sqrt{\frac{3.98^2}{20}+ \frac{4.03^2}{17}} = -1.63$

$(14.5-13.9) +1.69 \sqrt{\frac{3.98^2}{20}+ \frac{4.03^2}{17}} = 2.83$

And the confidence interval for this case $-1.63 \leq \mu_1 -\mu_2 \leq 2.83$

Step-by-step explanation:

We know the following info from the problem

$\bar X_1 = 14.5$ sample mean for the group 1

$s_1 = 3.98$ the standard deviation for the group 1

$n_1= 20$ the sample size for group 1

$\bar X_2 = 13.9$ sample mean for the group 2

$s_1 = 4.03$ the standard deviation for the group 2

$n_2= 17$ the sample size for group 2

We have all the conditions satisifed since we have random samples.

We want to construct a confidence interval for the true difference of means and the correct formula for this case is:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}$

The degrees of freedom are given :

$df = n_1 +n_2- 2 = 20+17-2=35$

The confidence level is 0.9 or 90% and the significance level is $\alpha=1-0.9=0.1$ and $\alpha/2 = 0.05$ and the critical value for this case is:

$t_{\alpha/2} = 1.69$

And replacing the info given we got:

$(14.5-13.9) -1.69 \sqrt{\frac{3.98^2}{20}+ \frac{4.03^2}{17}} = -1.63$

$(14.5-13.9) +1.69 \sqrt{\frac{3.98^2}{20}+ \frac{4.03^2}{17}} = 2.83$

And the confidence interval for this case $-1.63 \leq \mu_1 -\mu_2 \leq 2.83$

5 0

2 years ago

Solve the equation for<br> x 3/4(x+21) = 10.5 <br> x =

MrRa [10]

<h3>✍️ Answer:</h3><h3> $x=-\frac{294}{13}$ </h3>

✽✽✽✽

$\frac{x\cdot \:3}{4\left(x+21\right)}=10.5$

$\mathrm{Multiply\:both\:sides\:by\:}4\left(x+21\right)$

$\frac{x\cdot \:3}{4\left(x+21\right)}\cdot \:4\left(x+21\right)=10.5\cdot \:4\left(x+21\right)$

$3x=42\left(x+21\right)$

$3x=42x+882$

$\mathrm{Subtract\:}42x\mathrm{\:from\:both\:sides}$

$3x-42x=42x+882-42x$

$-39x=882$

$\mathrm{Divide\:both\:sides\:by\:}-39$

$\frac{-39x}{-39}=\frac{882}{-39}$

$x=-\frac{294}{13}$

❅❅❅❅❅❅❅❅❅❅

<h3>hope it helps...</h3><h3>have a great day!!</h3>

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vlada-n [284]

Answer:

(-4)^12

Step-by-step explanation:

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We see that vertical angles are congruent. So, angle 3 also = angle 5.

Alternate exterior angles are congruent.

So, angle 3 also = angle 1.

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Step-by-step explanation:

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