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Nadusha1986 [10]
3 years ago
10

88+88+88+88+88+88+88+88 times 88=_________

Mathematics
2 answers:
Svet_ta [14]3 years ago
8 0

Answer:

8,360

Step-by-step explanation:

88 + 88 + 88 + 88 + 88 + 88 + 88 + 88 × 88 is our problem.

Instead of adding all of these 88's together, we will just multiply 88 by 8 because there are 8 of these 88's.

That will leave us with: 704

But that is not our answer.

We still have one more step, and it is to multiply 704 by 8.

After we do that we will have: 8,360 as our answer.

I hope this helps! :>

Nastasia [14]3 years ago
3 0

Answer:

It's your friend at school. 8360 is your answer.

Step-by-step explanation:

88 + 88

= 176 + 88

= 264 + 88

= 352 + 88

= 440 + 88

= 528 + 88

= 616 + 88 =

704 x 88

= 8360

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Which equation could have been used to create this function table?    A.y = x + 9  B.y = x + 10  C.y = 9x  D.y = 10x
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The correct answer is a
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Brandon checked his bank account online and saw that he had a -$15 balance so he made a deposit.Now his balance is $67.Find d, t
kirill115 [55]

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82

Step-by-step explanation:

4 0
3 years ago
Can you guys please help me posted picture of question
Dimas [21]
Number of outcomes of rolling a die = 6
Number of outcomes from the spinner = 8

First we have to draw the 6 leaves to represent the 6 outcomes of rolling the die. Next on each of these leaves 8 possible outcomes of the spinner will be drawn using leaves.

So 6 leaves are being divided into further 8 leaves.

Thus total number of leaves will be  = 6 x 8 = 48 leaves

Therefore, the correct answer is option D 
3 0
3 years ago
. (0.5 point) We simulate the operations of a call center that opens from 8am to 6pm for 20 days. The daily average call waiting
SashulF [63]

Answer:

The 95% t-confidence interval for the difference in mean is approximately (-2.61, 1.16), therefore, there is not enough statistical evidence to show that there is a change in waiting time, therefore;

The change in the call waiting time is not statistically significant

Step-by-step explanation:

The given call waiting times are;

24.16, 20.17, 14.60, 19.79, 20.02, 14.60, 21.84, 21.45, 16.23, 19.60, 17.64, 16.53, 17.93, 22.81, 18.05, 16.36, 15.16, 19.24, 18.84, 20.77

19.81, 18.39, 24.34, 22.63, 20.20, 23.35, 16.21, 21.73, 17.18, 18.98, 19.35, 18.41, 20.57, 13.00, 17.25, 21.32, 23.29, 22.09, 12.88, 19.27

From the data we have;

The mean waiting time before the downsize, \overline x_1 = 18.7895

The mean waiting time before the downsize, s₁ = 2.705152

The sample size for the before the downsize, n₁ = 20

The mean waiting time after the downsize, \overline x_2 = 19.5125

The mean waiting time after the downsize, s₂ = 3.155945

The sample size for the after the downsize, n₂ = 20

The degrees of freedom, df = n₁ + n₂ - 2 = 20 + 20  - 2 = 38

df = 38

At 95% significance level, using a graphing calculator, we have; t_{\alpha /2} = ±2.026192

The t-confidence interval is given as follows;

\left (\bar{x}_{1}- \bar{x}_{2}  \right )\pm t_{\alpha /2}\sqrt{\dfrac{s_{1}^{2}}{n_{1}}+\dfrac{s_{2}^{2}}{n_{2}}}

Therefore;

\left (18.7895- 19.5152 \right )\pm 2.026192 \times \sqrt{\dfrac{2.705152^{2}}{20}+\dfrac{3.155945^2}{20}}

(18.7895 - 19.5125) - 2.026192*(2.705152²/20 + 3.155945²/20)^(0.5)

The 95% CI = -2.6063 < μ₂ - μ₁ < 1.16025996668

By approximation, we have;

The 95% CI = -2.61 < μ₂ - μ₁ < 1.16

Given that the 95% confidence interval ranges from a positive to a negative value, we are 95% sure that the confidence interval includes '0', therefore, there is sufficient evidence that there is no difference between the two means, and the change in call waiting time is not statistically significant.

6 0
2 years ago
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