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grin007 [14]
3 years ago
8

Please help need help asap!!

Mathematics
2 answers:
Ad libitum [116K]3 years ago
7 0
5v - 9 = 6
5v = 6 + 9
v = (6 + 9) / 5
v = 3


8v + 7 = 8 (3) + 7 = 31
VARVARA [1.3K]3 years ago
6 0
You can use Socratic for aleks it really helps :)
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Find the slope and use the point to write an equation of the line in point slope form.
Ainat [17]

Answer:

y - 3 = - 1/4(x + 2)

Step-by-step explanation:

m = - 1/4

Point = (-2,3)

y-intercept = 3 - (- 1/4) (-2)

= 3 - 1/2 = 5/2

y = - 1/4x + 5/2

7 0
3 years ago
Consists of the outcomes of 2 or more events
bagirrra123 [75]
Compound event is the answer 
3 0
4 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
Select all the expressions which are equivalent to –3(0.15 – 0.2 + 0.25p)
Sergio [31]

Answer:

–3(0.15 – 0.2 + 0.25p) = –3(0.15) + (–3)(0.2) + (–3)(0.25p)

Because the 2nd experssion is taking -3 times each number in the ( ).

hope this helps.

Step-by-step explanation:

4 0
3 years ago
Rewrite 2x^6-9x^5+4x^2-5/x^3-5 in the form q(x)+r(x)/b(x) what is q(x)?
WITCHER [35]
Hello,

(2x^6-9x^5+4x^2-5)|x^3-5
-(2x^6-10x^3 )         | 2x^3  -9x^2 +10
-------------------
-9x^5 +10x^3 +4x^2-5
-(-9x^5+45x^2)
---------------------------
10x^3-41x^2-5
-(10x^3-50)
-----------------
-41x²+45


Answer D

7 0
3 years ago
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