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3 years ago

Find the area of the semicircle

Mathematics

1 answer:

egoroff_w [7]3 years ago

5 0

Answer:

3.92yd

Step-by-step explanation:

pi r²÷2

pi 2.5²÷2=3.92

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The ratio of players to coaches at football camp was 15 to 2. If there were 120 players, how many

Marta_Voda [28]

Answer:

Step-by-step explanation:

120 divided by 15 is 8

8 x 2 = 16

There are 16 coaches.

8 0

3 years ago

I need helps ASapP please help look at pictures

vampirchik [111]

Answer:

I can help you but I'm not see it clear

7 0

3 years ago

Will 5x-3y>15 have a solid line

marysya [2.9K]

Answer:

The inequality $5x-3y >15$ have a dashed line

Step-by-step explanation:

we have

$5x-3y >15$

The solution of this inequality is the shaded area below the dashed line

The equation of the dashed line is $5x-3y=15$

The slope of the dashed line is positive

The y-intercept of the dashed line is -5 ---> point (0,-5)

The x-intercept of the dashed line is x=3 ----> point (3,0)

Graph the solution

see the attached figure

6 0

3 years ago

9. A large electronic office product contains 2000 electronic components. Assume that the probability that each component operat

Marysya12 [62]

Answer:

97.10% probability that five or more of the original 2000 components fail during the useful life of the product.

Step-by-step explanation:

For each component, there are only two possible outcomes. Either it works correctly, or it does not. The probability of a component falling is independent from other components. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem we have that:

$n = 2000, p = 1-0.995 = 0.005$

Approximate the probability that five or more of the original 2000 components fail during the useful life of the product.

We know that either less than five compoenents fail, or at least five do. The sum of the probabilities of these events is decimal 1. So

$P(X < 5) + P(X \geq 5) = 1$

We want $P(X \geq 5)$

$P(X \geq 5) = 1 - P(X < 5)$

In which

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{2000,0}.(0.005)^{0}.(0.995)^{2000} = 0.000044$

$P(X = 1) = C_{2000,1}.(0.005)^{1}.(0.995)^{1999} = 0.000445$

$P(X = 2) = C_{2000,2}.(0.005)^{2}.(0.995)^{1998} = 0.002235$

$P(X = 3) = C_{2000,3}.(0.005)^{3}.(0.995)^{1997} = 0.007480$

$P(X = 4) = C_{2000,4}.(0.005)^{4}.(0.995)^{1996} = 0.018765$

$P(X < 5) = P(X = 0) + `P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.000044 + 0.000445 + 0.002235 + 0.007480 + 0.018765 = 0.0290$

$P(X \geq 5) = 1 - P(X < 5) = 1 - 0.0290 = 0.9710$

97.10% probability that five or more of the original 2000 components fail during the useful life of the product.

4 0

3 years ago

Drag each value to the correct location on the table.

tatiyna

Answer:

Between 2 and 3: 3√101, 3√20, √7

Between 4 and 5: 3√90, √20

Between 9 and 10: √90, 3√800

3 0

2 years ago