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4 years ago

9

What’s the midpoint of (-2,4) and (7,3)

1 answer:

kirill115 [55]4 years ago

8 0

Answer:

the midpoint is (5/2, 7,2)

Step-by-step explanation:

You might be interested in

Which equation is in point-slope form and depicts the equation of this line?

Elodia [21]

We know
in the graph two points
point 1 (2,1) and point 2 (-1,-4)
the equation in point-slope form is ----------------> (y-y1)=m(x-x1)

step 1
find the value of m
m=(y2-y1)/(x2-x1)----------> (-4-1)/(-1-2)--------> m=5/3

step 2
with the point (-1,-4) and m=5/3
find the equation in point-slope form
(y-(-4))=(5/3)*(x-(-1))-------------> (y+4)=(5/3)(x+1)

the answer is the option d) ( y+ 4 ) =+ 5/3 ( x + 1 )

7 0

3 years ago

A distribution of values is normal with a mean of 220 and a standard deviation of 13. From this distribution, you are drawing sa

Paraphin [41]

Answer:

The interval containing the middle-most 48% of sample means is between 218.59 to 221.41.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are 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:

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

The Z-score measures how many standard deviations the measure 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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributied random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 220, \sigma = 13, n = 35, s = \frac{13}{\sqrt{35}} = 2.1974$

Find the interval containing the middle-most 48% of sample means:

50 - 48/2 = 26th percentile to 50 + 48/2 = 74th percentile. So

74th percentile

value of X when Z has a pvalue of 0.74. So X when Z = 0.643.

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

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$0.643 = \frac{X - 220}{2.1974}$

$X - 220 = 0.643*2.1974$

$X = 221.41$

26th percentile

Value of X when Z has a pvalue of 0.26. So X when Z = -0.643

$Z = \frac{X - \mu}{s}$

$-0.643 = \frac{X - 220}{2.1974}$

$X - 220 = -0.643*2.1974$

$X = 218.59$

The interval containing the middle-most 48% of sample means is between 218.59 to 221.41.

5 0

4 years ago

What is the 138-20+15

nlexa [21]

Then answer you seek is 133.

5 0

3 years ago

Suppose that two openings on an appellate court bench are to be filled from current municipal court judges. The municipal court

Ksju [112]

Answer:

$(a)\dfrac{92}{117}$

$(b)\dfrac{8}{39}$

$(c)\dfrac{25}{117}$

Step-by-step explanation:

Number of Men, n(M)=24

Number of Women, n(W)=3

Total Sample, n(S)=24+3=27

Since you cannot appoint the same person twice, the probabilities are <u>without replacement.</u>

(a)Probability that both appointees are men.

$P(MM)=\dfrac{24}{27}X \dfrac{23}{26}=\dfrac{552}{702}\\=\dfrac{92}{117}$

(b)Probability that one man and one woman are appointed.

To find the probability that one man and one woman are appointed, this could happen in two ways.

A man is appointed first and a woman is appointed next.
A woman is appointed first and a man is appointed next.

P(One man and one woman are appointed) $=P(MW)+P(WM)$

$=(\dfrac{24}{27}X \dfrac{3}{26})+(\dfrac{3}{27}X \dfrac{24}{26})\\=\dfrac{72}{702}+\dfrac{72}{702}\\=\dfrac{144}{702}\\=\dfrac{8}{39}$

(c)Probability that at least one woman is appointed.

The probability that at least one woman is appointed can occur in three ways.

A man is appointed first and a woman is appointed next.
A woman is appointed first and a man is appointed next.
Two women are appointed

P(at least one woman is appointed) $=P(MW)+P(WM)+P(WW)$

$P(WW)=\dfrac{3}{27}X \dfrac{2}{26}=\dfrac{6}{702}$

In Part B, $P(MW)+P(WM)=\frac{8}{39}$

Therefore:

$P(MW)+P(WM)+P(WW)=\dfrac{8}{39}+\dfrac{6}{702}\\$P(at least one woman is appointed)=\dfrac{25}{117}$

5 0

3 years ago

Let U={q, r, s, t, u, V, W, X, Y, Z},

trapecia [35]

Answer:

Ooh is theis unions I did it but

3 0

2 years ago