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alexira [117]
3 years ago
5

The weight of panda bears, P, is approximately Normally distributed with a mean of 185 pounds and a standard deviation of 12.3 p

ounds. The weight of koala bears, K, is approximately Normally distributed with a mean of 21.5 pounds and a standard deviation of 4.8 pounds. Suppose a zookeeper randomly chooses a panda and a koala bear, where P and K are independent random variables.
What is the probability that the total weight for the two animals is 225 pounds or more?

0.020
0.081
0.195
0.851
Mathematics
2 answers:
Xelga [282]3 years ago
4 0

Answer:

The CORRECT answer is B

Step-by-step explanation:

Dont mind the dork above me the correct answer is B on edge

katovenus [111]3 years ago
3 0

Answer:

c

Step-by-step explanation:

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A shopper bought the following items at the farmers market:
Montano1993 [528]
Ears 0.3 each
Apples .24 each
Tomatoes. 62
4 0
3 years ago
Suppose that four microchips in a production run of fifty are defective. A sample of six is to be selected to be checked for def
Irina-Kira [14]

Answer:

a) 15,890,700 different samples can be chosen

b) 6,523,881 will contain at least one defective chip

c) 41.1%

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

The order in which the microchips are chosen is not important. So we use the combinations formula to solve this question.

Combinations formula:

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)!}

(a) How many different samples can be chosen?

Samples of 6 from a set of 50.

Then

C_{50,6} = \frac{50!}{6!(50-6)!} = 15890700

15,890,700 different samples can be chosen

(b) How many samples will contain at least one defective chip?

Either a sample contains no defective chip, or it contains at least one. From a), the sum of them is 15890700.

Then

No defective chips:

Four are defective.

So 50-4 = 46 are not.

This is samples of 6 from a set of 46.

C_{46,6} = \frac{46!}{6!(46-6)!} = 9366819

9,366,819 samples contain no defective chips.

At least one:

9366819 + n = 15890700

n = 6523881

6,523,881 will contain at least one defective chip

(c) What is the probability (as a percent) that a randomly chosen sample of six contains at least one defective chip

6,523,881 out of 15,890,700

6,523,881/15,890,700 = 0.411

As a percent

41.1%

8 0
2 years ago
What is the sum of 2/5 and 3/4
kirill115 [55]
23/20 or 1 and 3/20
Hope this helped.
8 0
3 years ago
The value of n is both 5 times as much as the value of m and 36 more than the value of m. What are the values of n and m? Explai
mr Goodwill [35]
N = 5m
n = m + 36

Both equations are equal to n so they equal each other. Solve for m

5m = m + 36
5m - m= m + 36 - m
4m = 36
4m/4=36/4
m = 9

Solve for n by inputting value of m into the given equations.

n = 5m ........................ n = m + 36
n = 5 (9) ....................... n = 9 + 36
n = 45 ........................... n = 45

They match so it's Correct!

m = 9
n = 45
3 0
3 years ago
Jo's total charge for an onsite job is a fixed charge of $50 per job plus an hourly charge of $80.00 per hour.What would be the
Scrat [10]
X=hours worked and y=total
80x+50=y
6 0
3 years ago
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