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Helga [31]
3 years ago
7

What’s percent is equal to 7/20

Mathematics
2 answers:
gtnhenbr [62]3 years ago
5 0

Answer:

35%

Step-by-step explanation:

7 ÷ 20 = 0.35

0.35 is equal to 35%

BabaBlast [244]3 years ago
3 0

Answer:

The answer is 35%.

Step-by-step explanation:

This problem involves multiplication and division.

First, divide the numerator by the denominator (7 divided by 20).

You should get 0.35.

Next, multiply the quotient by 100 (0.35*100).

You should have 35.

So the answer is 35%.

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3 0
4 years ago
Read 2 more answers
Find the measure of the indicated angle to the nearest degree
Y_Kistochka [10]

Answer:

32°

Step-by-step explanation:

use this formula :

Now you want to get the angle be x, so the opposite side is 28 and hypotenuse is 53. Putting in the formula we get:

Sin(x) = 28/53

Sin(x)= 0.5283018868

So to find x just do Sin^-1

x = Sin^-1 (0.5283018868)

you get 31.8907918, ok so we just want the nearest degree and if we do that we get 32˚

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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].

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3 years ago
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erastova [34]
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On the interval [7π/4, 9π/4], the average slope is \dfrac{(\frac{\sqrt{2}}{2}-\frac{-\sqrt{2}}{2})}{(\frac{9\pi}{4}-\frac{7\pi}{4})}=\dfrac{2\sqrt{2}}{\pi}


The average rate of change is highest on the intervals [0, π/4] and [7π/4, 9π/4].
4 0
3 years ago
Please help me out!​
valina [46]

Answer:

4/11=.363636....

I hope this is good enough:

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