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Elan Coil [88]
2 years ago
13

What is 4 times 2. Will give brainliest HELPPPPPPPPPPPPPPPPPPPPPPPp

Mathematics
2 answers:
Temka [501]2 years ago
8 0

the answer is 8

Step-by-step explanation:

4 + 4=8

double the number

Alenkinab [10]2 years ago
7 0

Answer:

4 times 2= 8

Step-by-step explanation:

hope it helps

<em>mark me brainliest please</em>

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A spinner is divided into 8 equal sections, and each section contains a number from 1 to 8. What is the probability of the spinn
alukav5142 [94]

Answer:

more than likely none

Step-by-step explanation:

because the is no number higher than 8 that can go into 8 evenley

8 0
3 years ago
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A statistics textbook chapter contains 60 exercises, 6 of which are essay questions. A student is assigned 10 problems. (a) What
Scilla [17]

Answer:

a) P=0.3174

b) P=0.4232

c) P=0.2594

d) The shape of the hypergeometric, in this case, is like a binomial with mean np=1.

Step-by-step explanation:

The appropiate distribution to model this is the hypergeometric distribution:

P(X=x)=\frac{\binom{s}{x}\binom{N-s}{M-x}}{\binom{N}{M}}=\frac{\binom{6}{x}\binom{54}{10-x}}{\binom{60}{10}}

a) What is the probability that none of the questions are essay?

P(X=0)=\frac{\binom{6}{0}\binom{54}{10-0}}{\binom{60}{10}}\\\\P(X=0)=\frac{1*(54!/(10!*44!)}{60!/(10!*50!)} =\frac{2.3931*10^{10}}{7.5394*10^{10}} = 0.3174

b)  What is the probability that at least one is essay?

P(X=1)=\frac{\binom{6}{1}\binom{54}{9}}{\binom{60}{10}}\\\\P(X=1)=\frac{6*(54!/(9!*43!)}{60!/(10!*50!)} =\frac{3.1908*10^{10}}{7.5394*10^{10}} =0.4232

c) What is the probability that two or more are essay?

P(X\geq2)=1-(P(0)+P(1))=1-(0.3174+0.4232)=1-0.7406=0.2594

8 0
3 years ago
Recall that in a 30 – 60 – 90 triangle, if the shortest leg measures x units, then the longer leg measures xStartRoot 3 EndRoot
Over [174]

Answer:

150√3- 75π) ft². if you see it

Step-by-step explanation:

8 0
2 years ago
Please help! This is timed
gulaghasi [49]
I think the answer is C 1 2/3
4 0
3 years ago
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Lunch break: In a recent survey of 655 working Americans ages 25-34, the average weekly amount spent on lunch was $43.5
Inga [223]

Using the Empirical Rule, it is found that:

  • a) Approximately 99.7% of the amounts are between $35.26 and $51.88.
  • b) Approximately 95% of the amounts are between $38.03 and $49.11.
  • c) Approximately 68% of the amounts fall between $40.73 and $46.27.

------------

The Empirical Rule states that, in a <em>bell-shaped </em>distribution:

  • Approximately 68% of the measures are within 1 standard deviation of the mean.
  • Approximately 95% of the measures are within 2 standard deviations of the mean.
  • Approximately 99.7% of the measures are within 3 standard deviations of the mean.

-----------

Item a:

43.5 - 3(2.77) = 35.26

43.5 + 3(2.77) = 51.88

Within <em>3 standard deviations of the mean</em>, thus, approximately 99.7%.

-----------

Item b:

43.5 - 2(2.77) = 38.03

43.5 + 2(2.77) = 49.11

Within 2<em> standard deviations of the mean</em>, thus, approximately 95%.

-----------

Item c:

  • 68% is within 1 standard deviation of the mean, so:

43.5 - 2.77 = 40.73

43.5 + 2.77 = 46.27

Approximately 68% of the amounts fall between $40.73 and $46.27.

A similar problem is given at brainly.com/question/15967965

8 0
3 years ago
Read 2 more answers
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