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3 years ago

What number is halfway between 28 and 82

Mathematics

2 answers:

densk [106]3 years ago

6 0

Answer:

the answer is 55

Step-by-step explanation:

(28+82)/2

gtnhenbr [62]3 years ago

5 0

Answer:

Step-by-step explanation:

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The ratio of apples and oranges are 1:3 what would happen if I had 150 oranges

oksian1 [2.3K]

multiply the 3 by how ever much to get 150

there will be 50 apples

5 0

3 years ago

J.J.Bean sells a wide variety of outdoor equipment and clothing. The company sells both through mail order and via the internet.

melamori03 [73]

Answer:

99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

Step-by-step explanation:

We are given that a random sample of 16 sales receipts for mail-order sales results in a mean sale amount of $74.50 with a standard deviation of $17.25.

A random sample of 9 sales receipts for internet sales results in a mean sale amount of $84.40 with a standard deviation of $21.25.

The pivotal quantity that will be used for constructing 99% confidence interval for true mean difference is given by;

P.Q. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t__n_1_+_n_2_-_2$

where, $\bar X_1$ = sample mean for mail-order sales = $74.50

$\bar X_2$ = sample mean for internet sales = $84.40

$s_1$ = sample standard deviation for mail-order purchases = $17.25

$s_2$ = sample standard deviation for internet purchases = $21.25

$n_1$ = sample of sales receipts for mail-order purchases = 16

$n_2$ = sample of sales receipts for internet purchases = 9

Also, $s_p =\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(16-1)\times 17.25^{2}+(9-1)\times 21.25^{2} }{16+9-2} }$ = 18.74

The true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is represented by ( $\mu_1-\mu_2$ ).

Now, 99% confidence interval for ( $\mu_1-\mu_2$ ) is given by;

= $(\bar X_1-\bar X_2) \pm t_(_\frac{\alpha}{2}_) \times s_p \times \sqrt{\frac{1}{n_1} +\frac{1}{n_2}}$

Here, the critical value of t at 0.5% level of significance and 23 degrees of freedom is given as 2.807.

= $(74.50-84.40) \pm (2.807 \times 18.74 \times \sqrt{\frac{1}{16} +\frac{1}{9}})$

= [$-31.82 , $12.02]

Hence, 99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

5 0

3 years ago

Question in photo!! Will mark Brainlyist

victus00 [196]

Answer:

Option B.

$a$

Step-by-step explanation:

Given:

-3( -x + a ) + 7 < 5 When x = -1

To Find:

a ?

Solution:

-3( -x + a ) + 7 < 5 When x = -1

Substitute x = -1 in above we get

-3(-(-1) + a ) + 7 < 5

-3(1 + a) < 5 - 7

Using Distributive Property we get i.e A(B+C) = AB + AC

$\therefore -3\times 1 +(-3)\times a$

7 0

3 years ago

1 each of the two items that Val bought cost more than 10$ 2. Val spent 34$ for the 2 items. 3. Neither of the two items that Va

DerKrebs [107]

Statement : Val spent $ 34 for the 2 items

contradiction : Neither of the 2 items that Val bought cost more then $ 15

think about it....if neither of them cost more then 15....lets say they cost 15 each...then there is no way he spent $ 34.....he would spend at most $ 30

8 0

3 years ago

Read 2 more answers

Determine whether the integral is convergent or divergent. ∫[infinity] 2 e^−1/x / x^2 dx : O Convergent O divergent If it is con

monitta

Let $f(x)=e^{-1/x}$ . Then $f'(x)=\frac1{x^2}e^{-1/x}>0$ for all $x\ge2$ , so $f$ is strictly increasing. As $x\to\infty$ , $e^{-1/x}\to e^0=1$ , so $f$ is bounded above by 1. This is to say,