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cluponka [151]
3 years ago
11

What is the answer ?? Need help plis

Mathematics
2 answers:
bagirrra123 [75]3 years ago
8 0
Take a closer picture
Mashcka [7]3 years ago
5 0
It would be 8 minuts because i just know
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Lf oil costs $95 per barrel and $1 =R8. How much does a barrel cost in rands?.​
Elan Coil [88]

Answer:

R760

Step-by-step explanation:

If $1 is equivalent to R8 and the total oil costs $95 you must multiply 8x95. This leaves you with an answer of R760

7 0
3 years ago
What is 1/4 ➗ 9 in simplest form? and how do i simplify it?
iren [92.7K]

Answer:

1/36

Step-by-step explanation:

1/4 ÷ 9

= 1/4 x 1/9

= 1/36

6 0
3 years ago
Rewrite f(x) = 2(x − 1)2 + 3 from vertex form to standard form. (2 points)
Ostrovityanka [42]

Answer:

F(x)= 4x-1 I think so hope it will help you

7 0
3 years ago
To estimate the mean height μ of male students on your campus,you will measure an SRS of students. You know from government data
nexus9112 [7]

Answer:

a) \sigma = 0.167

b) We need a sample of at least 282 young men.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by

Z = \frac{X - \mu}{\sigma}

This Zscore is how many standard deviations the value of the measure X is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

(a) What standard deviation must x have so that 99.7% of allsamples give an x within one-half inch of μ?

To solve this problem, we use the 68-95-99.7 rule. This rule states that:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviations of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we want 99.7% of all samples give X within one-half inch of \mu. So X - \mu = 0.5 must have Z = 3 and X - \mu = -0.5 must have Z = -3.

So

Z = \frac{X - \mu}{\sigma}

3 = \frac{0.5}{\sigma}

3\sigma = 0.5

\sigma = \frac{0.5}{3}

\sigma = 0.167

(b) How large an SRS do you need to reduce the standard deviationof x to the value you found in part (a)?

You know from government data that heights of young men are approximately Normal with standard deviation about 2.8 inches. This means that \sigma = 2.8

The standard deviation of a sample of n young man is given by the following formula

s = \frac{\sigma}{\sqrt{n}}

We want to have s = 0.167

0.167 = \frac{2.8}{\sqrt{n}}

0.167\sqrt{n} = 2.8

\sqrt{n} = \frac{2.8}{0.167}

\sqrt{n} = 16.77

\sqrt{n}^{2} = 16.77^{2}

n = 281.23

We need a sample of at least 282 young men.

6 0
3 years ago
NEED HELP ASAP
mario62 [17]
1) (-7)3

2) PEMDAS; simplify (5-3)

3) 5^3, 5x5x5 and 125

4) -8^4

5) 36

6) 7

7) -100

8) 30

There ya go! Pretty sure they're all right!



5 0
3 years ago
Read 2 more answers
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