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2 years ago

The question is in the picture. (Alegbra 1)

Mathematics

2 answers:

morpeh [17]2 years ago

6 0

Answer:

Step-by-step explanation:

4/5a*3/2a^2

4*3/5a*2a^2

12/10a^3

6/5a^3

therefore ur answer is 6/5a^3

Varvara68 [4.7K]2 years ago

6 0

Answer:

option 3

Explanation :

$\frac{4}{5a} \cdot \frac{3}{2a^2} = \frac{12}{10 a^3} = \frac{6}{5a^3}$

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Step-by-step explanation:

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2 years ago

If an integer m is divisible by both 10 and 12, then it must also be divisible by which of the following?

Murrr4er [49]

Answer:

Step-by-step explanation:

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3 years ago

In a survey of adults who follow more than one sport, 30% listed football as their favorite sport. You survey 15 adults who foll

maxonik [38]

Answer: 0.5

Step-by-step explanation:

Binomial probability formula :-

$P(x)=^nC_x\ p^x(q)^{n-x}$ , where P(x) is the probability of getting success in x trials , n is the total trials and p is the probability of getting success in each trial.

Given : The probability that the adults follow more than one game = 0.30

Then , q= 1-p = 1-0.30=0.70

The number of adults surveyed : n= 15

Let X be represents the adults who follow more than one sport.

Then , the probability that fewer than 4 of them will say that football is their favorite sport,

$P(X\leq4)=P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)\\\\=^{15}C_{0}\ (0.30)^0(0.70)^{15}+^{15}C_{1}\ (0.30)^1(0.70)^{14}+^{15}C_{2}\ (0.30)^2(0.70)^{13}+^{15}C_{3}\ (0.30)^3(0.70)^{12}+^{15}C_{4}\ (0.30)^4(0.70)^{11}\\\\=(0.30)^0(0.70)^{15}+15(0.30)^1(0.70)^{14}+105(0.30)^2(0.70)^{13}+455(0.30)^3(0.70)^{12}+1365(0.30)^4(0.70)^{11}\\\\=0.515491059227\approx0.5$

Hence, the probability rounded to the nearest tenth that fewer than 4 of them will say that football is their favorite sport =0.5

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3 years ago

The figure below is divided into 100 small squares of equal size.

Bess [88]

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2 years ago

Read 2 more answers