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MissTica
3 years ago
12

A toy company intends to ship out its entire stock of 10,254 toy trucks, and three times as many toy cars. If each box holds 35

toys of either type, how many full boxes can be packed, and how many toys are left over for the last box?
Mathematics
1 answer:
svetoff [14.1K]3 years ago
6 0
There are 10,254 toy trucks

and 10,254 * 3 = 30,762 toy cars.

In total, there are 10,254+30,762= 41,016 toys 

when we divide 41,016 by 35, the quotient is the number of the full boxes, and the remainder is the number of toys left.

41,016=35*1171+31


Answer: 1171 full boxes, 31 toys left for the last box
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Answer and Explanation:

Given : Five males with an​ X-linked genetic disorder have one child each. The random variable x is the number of children among the five who inherit the​ X-linked genetic disorder.

To find :

a. Does the table show a probability distribution?

b. Find the mean and standard deviation of the random variable x.

Solution :

a) To determine that table shows a probability distribution we add up all six probabilities if the sum is 1 then it is a valid distribution.

\sum P(X)=0.029+0.147+0.324+0.324+0.147+0.029

\sum P(X)=1

Yes it is a probability distribution.

b) First we create the table as per requirements,

x      P(x)         xP(x)           x²            x²P(x)

0    0.029         0              0                0

1     0.147        0.147           1              0.147

2    0.324       0.648         4              1.296

3    0.324       0.972         9              2.916

4    0.147        0.588        16              2.352

5    0.029       0.145         25            0.725

   ∑P(x)=1      ∑xP(x)=2.5               ∑x²P(x)=7.436

The mean of the random variable is

\mu=\sum xP(x)=2.5

The standard deviation of the random sample is

Variance=\sigma^2

\sigma^2=\sum x^2P(x)-\mu^2

\sigma^2=7.436-(2.5)^2

\sigma^2=7.436-6.25

\sigma^2=1.186

\sigma=\sqrt{1.186}

\sigma=1.08

Therefore, The mean is 2.5 and the standard deviation is 1.08.

5 0
3 years ago
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