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3 years ago

#13 I'm not sure how to do it ;~; I need help pls

Mathematics

1 answer:

DerKrebs [107]3 years ago

4 0

You have climbed 3/4 of the wall and you are 24 feet above the ground. The best equation I can think 24/3 + 24 = h.

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Brainliest for right answer?

shusha [124]

Step-by-step explanation:

total sold = 136

hot dogs sold : soda - 30

136 = soda + hot dogs

136 = soda + (soda -30)

136 = 2soda -30

2soda = 166

soda sold = 83

hot dogs sold = 83-30 = 53

6 0

2 years ago

Read 2 more answers

Help I will mark brainliestttt

Fynjy0 [20]

<h2>its D because u can </h2><h2> and add / mutiply *</h2><h2> </h2>

8 0

3 years ago

Read 2 more answers

The steps below show the incomplete solution to find the value of x for the equation

monitta

The next step is most likely to add 3 to both sides, so that 3 cancels out, and 15 turns into 18.

The answer is: Step 4: 3x = 18
(I don't know if you want me to continue or not, but I will) ;)

Then divide by 3 on both sides so that 3 cancels out and 18 turns into 6, and you have the value of x.

Step 5: x = 6

You can check your answer by replacing x with 6.

5 · 6 - 2 · 6 - 3 = -2 + 17
15 = 15

Hope this helped! If you have anymore questions or don't understand, please comment or DM me. :)

5 0

3 years ago

What are the domain and range of the function f(x)=3/4x+5?

dybincka [34]

Start with parentheses

7 0

3 years ago

Land's Bend sells a wide variety of outdoor equipment and clothing. The company sells both through mail order and via the intern

Naya [18.7K]

Answer:

80% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases ( $\mu_1-\mu_2$ ) is [-9.132 , 23.332].

Step-by-step explanation:

We are given that a random sample of 7 sales receipts for mail-order sales results in a mean sale amount of $81.70 with a standard deviation of $18.75.

A random sample of 11 sales receipts for internet sales results in a mean sale amount of $74.60 with a standard deviation of $28.25.

Firstly, the Pivotal quantity for 80% confidence interval for the difference between population means is given by;

P.Q. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ~ $t__n__1-_n__2-2$

where, $\bar X_1$ = sample mean sales receipts for mail-order sales = $81.70

$\bar X_2$ = sample mean sales receipts for internet sales = $74.60

$s_1$ = sample standard deviation for mail-order sales = $18.75

$s_2$ = sample standard deviation for internet sales = $28.25

$n_1$ = size of sales receipts for mail-order sales = 7

$n_2$ = size of sales receipts for internet sales = 11

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2} +(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(7-1)\times 18.75^{2} +(11-1)\times 28.25^{2} }{7+11-2} }$ = 25.11

<em>Here for constructing 80% confidence interval we have used Two-sample t test statistics as we don't know about population standard deviations.</em>

<em />

So, 80% confidence interval for the difference between population means, ( $\mu_1-\mu_2$ ) is ;

P(-1.337 < $t_1_6$ < 1.337) = 0.80 {As the critical value of t at 16 degree

of freedom are -1.337 & 1.337 with P = 10%}

P(-1.337 < $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < 1.337) = 0.80

P( $-1.337 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < ${(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}$ < $1.337 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ) = 0.80

P( $(\bar X_1-\bar X_2)-1.337 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < ( $\mu_1-\mu_2$ ) < $(\bar X_1-\bar X_2)+1.337 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ) = 0.80

<u>80% confidence interval for</u> ( $\mu_1-\mu_2$ ) =

[ $(\bar X_1-\bar X_2)-1.337 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ , $(\bar X_1-\bar X_2)+1.337 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ]

= [ $(81.70-74.60)-1.337 \times {25.11 \times \sqrt{\frac{1}{7} +\frac{1}{11} } }$ , $(81.70-74.60)+1.337 \times {25.11 \times \sqrt{\frac{1}{7} +\frac{1}{11} } }$ ]

= [-9.132 , 23.332]

Therefore, 80% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases ( $\mu_1-\mu_2$ ) is [-9.132 , 23.332].

4 0

3 years ago