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3 years ago

Giving 100 points if someone can answer this correctly...

Mathematics

1 answer:

RUDIKE [14]3 years ago

7 0

Answer:

3813.6

Step-by-step explanation:

90.8 x 42 = 3813.6

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I rly need help with this..plz helpp

k0ka [10]

Probably let x be 8 and y be 2
Then you input the values
8-2=6
Then
2(8)-10(2)
=16-20
=-4

How they have 4?!

5 0

2 years ago

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Moderators<br><br><br>Need answers

GarryVolchara [31]

Step-by-step explanation:

please mark me as brainlest

4 0

2 years ago

The SAT scores have an average of 1200 with a standard deviation of 60. A sample of 36 scores is selected. a) What is the probab

erastova [34]

Answer:

the probability that the sample mean will be larger than 1224 is 0.0082

Step-by-step explanation:

Given that:

The SAT scores have an average of 1200

with a standard deviation of 60

also; a sample of 36 scores is selected

The objective is to determine the probability that the sample mean will be larger than 1224

Assuming X to be the random variable that represents the SAT score of each student.

This implies that ;

$S \sim N ( 1200,60)$

the probability that the sample mean will be larger than 1224 will now be:

$P(\overline X > 1224) = P(\dfrac{\overline X - \mu }{\dfrac{\sigma}{\sqrt{n}} }> \dfrac{}{}\dfrac{1224- \mu }{\dfrac{\sigma}{\sqrt{n}} })$

$P(\overline X > 1224) = P(Z > \dfrac{1224- 1200 }{\dfrac{60}{\sqrt{36}} })$

$P(\overline X > 1224) = P(Z > \dfrac{24 }{\dfrac{60}{6} })$

$P(\overline X > 1224) = P(Z > \dfrac{24 }{10} })$

$P(\overline X > 1224) = P(Z > 2.4 })$

$P(\overline X > 1224) =1 - P(Z \leq 2.4 })$

From Excel Table ; Using the formula (=NORMDIST(2.4))

P(\overline X > 1224) = 1 - 0.9918

P(\overline X > 1224) = 0.0082

Hence; the probability that the sample mean will be larger than 1224 is 0.0082

4 0

3 years ago

How does the number 8 compare to number of 80

olga2289 [7]

Answer:

8 times 10 is 80

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

What is the average value of latex: y=\tan\left(\frac{x^2}{9}\right) y = tan ⁡ ( x 2 9 ) on the closed interval [1.25, 2]?

german

For this case we have the following function:
$f(x) = tan(\frac{x^2}{9})$
The average rate of change is given by:
$AVR = \frac{f(x2) - f(x1)}{x2- x1}$
Evaluating the function for the given interval we have:
For x = 1.25:
$f(1.25) = tan(\frac{1.25^2}{9})$
$f(1.25) = 0.18$
For x = 2:
$f(2) = tan(\frac{2^2}{9})$
$f(2) = 0.48$
Then, replacing values we have:
$AVR = \frac{0.48 - 0.18}{2 - 1.25}$
$AVR = 0.4$
Answer:
the average value of on the closed interval [1.25, 2] is:
$AVR = 0.4$

3 0

3 years ago