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3 years ago

Please help me with this equation! you will be marked as brainliest

Mathematics

2 answers:

notka56 [123]3 years ago

7 0

Answer:

option c is write

i make sure this answer is true

lesya692 [45]3 years ago

4 0

The answer is C in a simplified term.

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Order each set of numbers from LEAST to GREATEST.<br><br> -4 , -5/4 , 0 , -5, -1.5

frosja888 [35]

-5, -4, -1.5, -5/4, 0

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3 years ago

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In Sam’s pet corner, there are puppies and parrots, the total of 15 heads and 50 feet. How many parrots and how many puppies?

laila [671]

Answer:

Since the number of feet = 50

We can start by taking the number of puppies as 10(Trial and Error Method)

Since puppies have 4 feet,

10 puppies will have 40 feet.

Parrots have 2 feet so the remaining pets, i.e., 5 parrots will have 10 feet.

Therefore the total = 50 feet

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4 years ago

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3 years ago

Olympic gold medal winner Michael Phelps compete in a pool with required dimensions 2.5 M by 50 m by 2 m what is the volume of t

Murrr4er [49]

Answer:

250 cubic m

Step-by-step explanation:

volume of the Olympic grid pool

$= 2.5 \times 50 \times 2 \\ = 2.5 \times 100 \\ = 250 \: {m}^{3}$

5 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