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Vilka [71]
3 years ago
9

Please help its due today WILL GIVE BRAINLIEST :)

Mathematics
1 answer:
ira [324]3 years ago
6 0
To find length and width you need to factor 3x² - 10x - 8. The final factored form will be (3x+2)(x-4).

Work:

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The cargo space of the box is 5.6 ft wide, 4.2 ft high and 3.8 ft long. How many cubic feet of cargo space does the box have?​
Romashka-Z-Leto [24]
The box has 89.38 cubic feet
8 0
3 years ago
The age of the children in kindergarten on the first day of school is uniformly distributed between 4.8 and 5.8 years old. A fir
Kazeer [188]

Answer:

(1) (c) <u>5.30 years</u>.

(2) (b) <u>0.289</u>.

(3) (b) <u>0.80</u>.

(4) (d) <u>0.50</u>.

(5) (a) <u>5.25 years</u>.

Step-by-step explanation:

Let <em>X</em> = age of the children in kindergarten on the first day of school.

The random variable <em>X</em> follows a continuous Uniform distribution with parameters <em>a</em> = 4.8 years and <em>b</em> = 5.8 years.

The probability density function function of <em>X</em> is:

f_{X}(x)=\left \{ {{\frac{1}{b-a}} ;\ a

(1)

The expected value of a Uniform random variable is:

E(X)=\frac{1}{2}(a+b)

Compute the mean of <em>X</em> as follows:

E(X)=\frac{1}{2}(a+b)=\frac{1}{2}\times (4.8+5.8)=5.3

Thus, the  mean of the distribution is (c) <u>5.30 years</u>.

(2)

The standard deviation of a Uniform random variable is:

SD(X)=\sqrt{\frac{1}{12}(b-a)^{2}}

Compute the standard deviation of <em>X</em> as follows:

SD(X)=\sqrt{\frac{1}{12}(b-a)^{2}}=\sqrt{\frac{1}{12}\times (5.8-4.8)^{2}}=0.289

Thus, the standard deviation of the distribution is (b) <u>0.289</u>.

(3)

Compute the probability that a randomly selected child is older than 5 years old as follows:

P(X>5)=\int\limits^{5.8}_{5} {\frac{1}{5.8-4.8}}\, dx\\

                =\int\limits^{5.8}_{5} {1}\, dx\\=[x]^{5.8}_{5}\\=(5.8-5)\\=0.8

Thus, the probability that a randomly selected child is older than 5 years old is (b) <u>0.80</u>.

(4)

Compute the probability that a randomly selected child is between 5.2 years and 5.7 years old as follows:

P(5.2

                            =\int\limits^{5.7}_{5.2} {1}\, dx\\=[x]^{5.7}_{5.2}\\=(5.7-5.2)\\=0.5

Thus, the probability that a randomly selected child is between 5.2 years and 5.7 years old is (d) <u>0.50</u>.

(5)

It is provided that a randomly selected child is at the 45th percentile.

This implies that:

P (X < x) = 0.45

Compute the value of <em>x</em> as follows:

   P (X < x) = 0.45

\int\limits^{x}_{4.8} {\frac{1}{5.8-4.8}}\, dx=0.45

        \int\limits^{x}_{4.8} {1}\, dx=0.45

           [x]^{x}_{4.8}=0.45

       x-4.8=0.45\\

                x=0.45+4.8\\x=5.25

Thus, the age of the child at the 45th percentile is (a) <u>5.25 years</u>.

6 0
3 years ago
Question 6 of 15, Step 1 of 1
OLga [1]

Answer:

3 hours

Step-by-step explanation:

369 divided by (58+65)

8 0
3 years ago
A bag contains four red marble, five blue marbles, and two green marbles. if one marble is chosen randomly, what is the probabil
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4/11 
Hope this helps :)

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Smallest to largest. Here are four decimal numbers: 0.9, 0.5, 0.8, 0.1
Pani-rosa [81]

Answer:

0.1,0.5,0.8,0.9

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
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