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3 years ago

Brainers, please help me, please.

Mathematics

1 answer:

Bogdan [553]3 years ago

7 0

Answer:

772.2

Step-by-step explanation:

Using the method of cross- product

• $\frac{a}{b}$ = $\frac{c}{d}$ ⇒ ad = bc, thus

85.8 × 9 = 11 × 70.2 = 772.2

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Step-by-step explanation:

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3 years ago

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jolli1 [7]

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PLEASE HELP FAST!!!!!!!!

seropon [69]

Answer:

Option B. -j = -h/-k is not correct

Step-by-step explanation:

As from the given scenario both the negative signs will be cancelled out giving positive j : -h/-k = -j

First option has: -j = -h/k In this case also the negative sign from both sides would be cancelled out.

Second option has: -h/-k = -j In this case negative signs cannot be cancelled out.

Third option has: h/-k = -j , negative sign would be cancelled from both sides.

Fourth option has: h/k = j , no negative sign on either side.

i hope it will help you!

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3 years ago

Find the slope of (6,2) (8,4)

denis23 [38]

Answer:

2/2 = 1

Step-by-step explanation:

y2 - y1 4 - 2 2 Equals 1

x2 - x1 8 - 6 2

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3 years ago

Assume the population has a normal distribution. A sample of 25 randomly selected students Has a mean test score of 81.5 With a

Natasha_Volkova [10]

Answer:

The 90% confidence interval for the mean test score is between 77.29 and 85.71.

Step-by-step explanation:

We have the standard deviation for the sample, so we use the t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 25 - 1 = 24

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 24 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.9}{2} = 0.95$ . So we have T = 2.064

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 2.064\frac{10.2}{\sqrt{25}} = 4.21$

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 81.5 - 4.21 = 77.29

The upper end of the interval is the sample mean added to M. So it is 81.5 + 4.21 = 85.71.

The 90% confidence interval for the mean test score is between 77.29 and 85.71.

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3 years ago