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3 years ago

Plzz help with this question

Mathematics

2 answers:

alexandr402 [8]3 years ago

7 0

Answer:

rip Pop we aint neva gettin cool with the opps

190000

hope this helps

Step-by-step explanation:

Reika [66]3 years ago

7 0

1230 is the answer to the question

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HELP!! 15 Points!!

kow [346]

Answer:

38.3 square inches

Step-by-step explanation:

The product of length and width gives the area of a rectangle:

area = (9.00 in)·(4.25 in) = 38.2500 in²

Each contributor has 3 significant digits, so your rounding rule requires the answer be rounded to 3 significant digits:

area ≈ 38.3 in²

3 0

3 years ago

Between which two rational numbers is √3 located? A.1.2 and 1.3 B.1.7 and 1.8 C.2.6 and 2.7 D. 2.7 and 2.8

julsineya [31]

B is the correct anwser

7 0

3 years ago

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I don’t under this .

melomori [17]

Answer:

The answer is QR,RS, and ST

Step-by-step explanation:

i hope this helps .

6 0

3 years ago

Read 2 more answers

Here is a sequence of numbers:

Artemon [7]

The first number is 34 and the second number is 63

7 0

3 years ago

A scientist inoculates mice, one at a time, with a disease germ until he finds 3 that have contracted the disease. If the prob

UkoKoshka [18]

Answer:

The probability that 8 mice are required is 0.2428.

Step-by-step explanation:

Given : A scientist inoculates mice, one at a time, with a disease germ until he finds 3 that have contracted the disease. If the probability of contracting the disease is two sevenths.

To find : What is the probability that 8 mice are required? The probability that that 8 mice are required is nothing ?

Solution :

Applying binomial distribution,

$P(X=r)=^nC_r p^rq^{n-r}$

Where, p is the probability of success $p=\frac{2}{7}$

q is the probability of failure q=1-p, $q=1-\frac{2}{7}=\frac{5}{7}$

n is total number of trials n=8

r=3

Substitute the values,

$P(X=3)=^8C_3 (\frac{2}{7})^3 (\frac{5}{7})^{8-3}$

$P(X=3)=\frac{8!}{3!5!}\times \frac{8}{343}\times (\frac{5}{7})^{5}$

$P(X=3)=\frac{8\times 7\times 6}{3\times 2\times 1}\times \frac{8}{343}\times \frac{3125}{16807}$

$P(X=3)=0.2428$

Therefore, the probability that 8 mice are required is 0.2428.

5 0

4 years ago