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sdas [7]
3 years ago
13

A football player completes a pass 69.4​% of the time. Find the probability that​ (a) the first pass he completes is the second​

pass, (b) the first pass he completes is the first or second​ pass, and​ (c) he does not complete his first two passes.
Mathematics
1 answer:
wlad13 [49]3 years ago
6 0

Step-by-step explanation:

(a) If his second pass is the first that he completes, that means he doesn't complete his first pass.

P = P(not first) × P(second)

P = (1 − 0.694) (0.694)

P ≈ 0.212

(b) This time we're looking for the probability that he doesn't complete the first but does complete the second, or completes the first and not the second.

P = P(not first) × P(second) + P(first) × P(not second)

P = (1 − 0.694) (0.694) + (0.694) (1 − 0.694)

P ≈ 0.425

(c) Finally, we want the probability he doesn't complete either pass.

P = P(not first) × P(not second)

P = (1 − 0.694) (1 − 0.694)

P ≈ 0.094

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Please show working out thanks
IgorLugansk [536]

Answer:

10m x 15m

Step-by-step explanation:

You are given some information.

1. The area of the garden: A₁ = 150m²

2. The area of the path: A₂ = 186m²

3. The width of the path: 3m

If the garden has width w and length l, the area of the garden is:

(1) A₁ = l * w

The area of the path is given by:

(2) A₂ = 3l + 3l + 3w + 3w + 4*3*3 = 6l + 6w + 36

Multiplying (2) with l gives:

(3) A₂l = 6l² + 6lw + 36l

Replacing l*w in (3) with A₁ from (1):

(4) A₂l = 6l² + 6A₁ + 36l

Combining:

(5) 6l² + (36 - A₂)l +6A₁ = 0

Simplifying:

(6) l² - 25l + 150 = 0

This equation can be factored:

(7) (l - 10)*(l - 15) = 0

Solving for l we get 2 solutions:

l₁ = 10, l₂ = 15

Using (1) to find w:

w₁ = 15, w₂ = 10

The two solutions are equivalent. The garden has dimensions 10m and 15m.

3 0
3 years ago
Pleaee help me with this im stuck
AURORKA [14]

the answers are clear if you use priority and equation rules.

  1. q=11
  2. u=2
  3. a=13
  4. l=8
  5. s=12
6 0
3 years ago
Read 2 more answers
Andrew uses 630 one-inch unit cubes to completely fill the inside of a rectangular box. Which could be the dimensions of the box
dsp73

we have 630 one-inch unit cubes and we want to completely fill the rectangular box (unknown dimensions).

If all the cubes are fitted tightly inside rectangular box without living any space, then box volume would be equal to cubes volume.

There are 630 one-inch unit cubes, so volume of cubes = 630 cubic inches.

Now the volume of rectangular box would also be 630 cubic inches.

We know the formula for volume of rectangular box = length × width × height.

So we need to find any three positive integers whose product is 630.

Out of all given choices, only option A satisfies the condition of factors of 630.

Hence, option A i.e. (7 in x 9 in x 10 in) is the final answer.

7 0
3 years ago
Find the value of x and y.
netineya [11]
Can you provide more info?
3 0
3 years ago
An electronic device factory is studying the length of life of the electronic components they produced. The manager selects two
Temka [501]

Answer:

The confidence interval will be given by:

200000 \pm 44.72z, in which z is related to the confidence level.

For a confidence level of x%, z is the value in the z-table that has a pvalue of 1 - \frac{1 - z}{2}

Step-by-step explanation:

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In the context of this problems:

It means that the sampling distributions of the sample mean of 500 components will be approximated normal, with mean 200,000 and standard deviation s = \frac{1000}{\sqrt{500}} = 44.72

To build the confidence interval:

The confidence interval for the average length of life of the electronic components they produced will be given by:

200000 \pm 44.72z, in which z is related to the confidence level.

For a confidence level of x%, z is the value in the z-table that has a pvalue of 1 - \frac{1 - z}{2}

3 0
3 years ago
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