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Maurinko [17]
3 years ago
6

Will give brainliest and extra points if help

Mathematics
2 answers:
stira [4]3 years ago
6 0

Answer: c

Step-by-step explanation:

vovikov84 [41]3 years ago
3 0

Answer:

I'm pretty sure it's B

Step-by-step explanation:

hope I helped

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B - (- 7) =-1<br> what is b?
irina1246 [14]

B-(-7)= -1

First negative multiply by negative is POSITIVE.

So: B+7= -1

Subtract 7 from both sides

B= -7-1

B= -8.

Answer: -8.

To Check Answer: Plug in the variable of B in the original equation.

So: -8-(-7)= -1

-8+7= -1

-1= -1

If the Solution works out the answer is right.

4 0
3 years ago
Read 2 more answers
Rename 1/2 and 1/5 using the least common denominator.
gladu [14]
C multiply 1/2 by 10 and 1/5 by 2
3 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
What is 10 times as many as 100
vampirchik [111]
10 \times 10 =100
7 0
3 years ago
0.1 is a rational number because it can be written as
Gnesinka [82]

Answer:

because it can be written in the <u>form</u><u> </u>of p/q

Where p and q are integers, and q is <u>not</u> equal to zero.

8 0
2 years ago
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