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4 years ago

6

What is

B20%20%20%5Cdiv%20%7D%2052%7D%20" id="TexFormula1" title="87251 {2}^{9} 3 \times \frac{23}{25 \sqrt[5]{20 \div } 52} " alt="87251 {2}^{9} 3 \times \frac{23}{25 \sqrt[5]{20 \div } 52} " align="absmiddle" class="latex-formula">

1 answer:

anygoal [31]4 years ago

8 0

Answer:

the answer is in picture...

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I shared 3 cars with my cousins I have 5 cousins and me. I want to know how many people will be in each car

tatuchka [14]

there will be 2 people in each car

5 0

3 years ago

A contractor is required by a county planning department to submit one, two, three, four, or five forms (depending on the nature

Westkost [7]

Answer:

(a) The value of <em>k</em> is $\frac{1}{15}$ .

(b) The probability that at most three forms are required is 0.40.

(c) The probability that between two and four forms (inclusive) are required is 0.60.

(d) $P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5$ is not the pmf of <em>y</em>.

Step-by-step explanation:

The random variable <em>Y</em> is defined as the number of forms required of the next applicant.

The probability mass function is defined as:

$P(y) = \left \{ {{ky};\ for \ y=1,2,...5 \atop {0};\ otherwise} \right$

(a)

The sum of all probabilities of an event is 1.

Use this law to compute the value of <em>k</em>.

$\sum P(y) = 1\\k+2k+3k+4k+5k=1\\15k=1\\k=\frac{1}{15}$

Thus, the value of <em>k</em> is $\frac{1}{15}$ .

(b)

Compute the value of P (Y ≤ 3) as follows:

$P(Y\leq 3)=P(Y=1)+P(Y=2)+P(Y=3)\\=\frac{1}{15}+\frac{2}{15}+ \frac{3}{15}\\=\frac{1+2+3}{15}\\ =\frac{6}{15} \\=0.40$

Thus, the probability that at most three forms are required is 0.40.

(c)

Compute the value of P (2 ≤ Y ≤ 4) as follows:

$P(2\leq Y\leq 4)=P(Y=2)+P(Y=3)+P(Y=4)\\=\frac{2}{15}+\frac{3}{15}+\frac{4}{15}\\ =\frac{2+3+4}{15}\\ =\frac{9}{15} \\=0.60$

Thus, the probability that between two and four forms (inclusive) are required is 0.60.

(d)

Now, for $P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5$ to be the pmf of Y it has to satisfy the conditions:

$P(y)=\frac{y^{2}}{50}>0;\ for\ all\ values\ of\ y \\$
$\sum P(y)=1$

<u>Check condition 1:</u>

$y=1:\ P(y)=\frac{y^{2}}{50}=\frac{1}{50}=0.02>0\\y=2:\ P(y)=\frac{y^{2}}{50}=\frac{4}{50}=0.08>0 \\y=3:\ P(y)=\frac{y^{2}}{50}=\frac{9}{50}=0.18>0\\y=4:\ P(y)=\frac{y^{2}}{50}=\frac{16}{50}=0.32>0 \\y=5:\ P(y)=\frac{y^{2}}{50}=\frac{25}{50}=0.50>0$

Condition 1 is fulfilled.

<u>Check condition 2:</u>

$\sum P(y)=0.02+0.08+0.18+0.32+0.50=1.1>1$

Condition 2 is not satisfied.

Thus, $P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5$ is not the pmf of <em>y</em>.

7 0

3 years ago

How can I solve this

Gennadij [26K]

24 ^ 2/3 = (8 ^ 2/3) * (3 ^ 2/3)

cancle out the 8^2/3 and you get (3 ^ 2/3) * (3 ^ 4/3) = 3^2 = 9

9

6 0

3 years ago

( 3 x 2 ) ^ 4 x 3 ^ -3 / 2 ^ 3 x 3 =<br><br> Simplify the exponential expression

Korvikt [17]

Answer:

2

Step-by-step explanation:

$\frac{(3*2)^4*3^{-3}}{2^3*3}$

we start simplifying by removing the parenthesis

Multiply the exponents inside the the parenthesis

3^4 * 2^4

$\frac{3^4*2^4*3^{-3}}{2^3*3}$

Now we apply exponential property

a^m * a^n = a^ (m+n)

3^4 * 3^-3 = 3^ (4-3) = 3^1

3 or 3^1 are same

$\frac{3^1*2^4}{2^3*3^1}$

3^1 at the top and bottom so we cancel it out

\frac{2^4}{2^3}

we apply log property . a^m / a^n = a^m-n

Now subtract the exponents

2^(4-3) = 2^1 = 2

4 0

4 years ago

A Principal needs 2 chaperones for every 25 students at the school dance how many chaperones does he need if 225 students are ex

Makovka662 [10]

Answer:

18 chaperones

Step-by-step explanation:

1. How many times does 25 go into 225

2. Whatever that number is multiply it by two

3. That is how you get the number of chaperones

4. THE END

5 0

4 years ago