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Klio2033 [76]
3 years ago
7

Simplify (7)(-6). a) -42 b) -13 c) 1 d)42

Mathematics
2 answers:
SOVA2 [1]3 years ago
7 0

-42 because 7 times -6 will equal -42 not positive 42

rusak2 [61]3 years ago
6 0
It’s A) simplify it you’ll get -42 because a negative times a positive is an negative
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The slope is 4. That eliminates the first two answers

Y-y1=m(x-x1)
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In 1985, the average ACT score was 18 with a standard deviation of 6. Also in 1985, the average SAT score was
lord [1]

The ACT score was better since the z-score of the ACT is greater than the z-score of the SAT. Then the correct option is B.

<h3>What is a normal distribution?</h3>

It is also called the Gaussian Distribution. It is the most important continuous probability distribution. The curve looks like a bell, so it is also called a bell curve.

The z-score is a numerical measurement used in statistics of the value's relationship to the mean of a group of values, measured in terms of standards from the mean.

The z-score is given as

\rm z = \dfrac{x - \mu }{\sigma}

In 1985, the average ACT score was 18 with a standard deviation of 6.

Also in 1985, the average SAT score was 500 with a standard deviation of 100.

Dr. Robertson took both tests in 1985 and scored a 26 on the ACT and a 620 on the SAT.

Then the z-score of the ACT will be

\rm z = \dfrac{26 - 18}{6}\\\\z = 1.3333

Then the z-score of the SAT will be

\rm z = \dfrac{620 - 500}{100}\\\\z = 1.2

Then the ACT score was better since the z-score of the ACT is greater than the z-score of the SAT.

More about the normal distribution link is given below.

brainly.com/question/12421652

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6 0
2 years ago
A coin is tossed 10 times resulting in 7 heads and 3 tails. The same coin is tossed 1000 times resulting in 510 heads and 490 ta
KiRa [710]
The answer is <span>C. 50%.

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8 0
3 years ago
Suppose that we have a sample space S5 {E1, E2, E3, E4E4E, E5, E6, E7}, where E1, E2, . . ., E7 denote the sample points. The fo
VashaNatasha [74]

Answer:

P(A) = 0.4 ; P(B) = 0.50 ; P(C) = 0.60 ; P(A u B) = 0.65 ; P(A n B) = 0.25 ;

A and C are mutually exclusive ;

0.5

Step-by-step explanation:

S = {E1, E2, E3, E4, E5, E6, E7}

P(E1) = .05, P(E2) = .20, P(E3) = .20, P(E4) = .25, P(E5) = .15, P(E6) = .10, and P(E7) = .05

Let:

A{E1,E4,E6}

B{E2,E4,E7}

C{E2,E3,E5,E7}

P(A) = 0.05 + 0.25 + 0.10 = 0.40

P(B) = 0.20 + 0.25 + 0.05 = 0.50

P(C) = 0.20 + 0.20 + 0.15 + 0.05 = 0.60

AUB = {E1, E2, E4, E6, E7}

P(A u B) = 0.05 + 0.20 + 0.25 + 0.10 + 0.05 = 0.65

AnB = {E4}

P(A n B) = 0.25

A and C are mutually exclusive if AnC = ∅

A n C = ∅

Hence, A and C are mutually exclusive.

B complement = B' = {E1, E3, E5, E6}

P(B') = 0.05 + 0.20 + 0.15 + 0.10 = 0.5

5 0
3 years ago
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