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3 years ago

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1 answer:

4 0

<span>From the problem, you can get this equation:
Jane is 2 years older than Sally
J=z+2
David is 4 times older than Jane.
D=4J

</span>Write two equivalent expressions that represent David’s age in terms of Sally’s age, z, and explain how the two expressions relate to one another.<span>Sally is z years old.
From the 2 equation above, you can get:
</span>D=4J
D=4(z+2)
D=4z+8

D-8=4z
(D-8)/4=z
The two equivalent expression would be D=4z+8 and (D-8)/4=z

The expression that more usefull would be D=4z+8 if you want to determine David age.

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What is the value of tan A?

Ronch [10]

For the given triangle, the tan of angle A equals $\frac{3}{4} = 0.75.$

Step-by-step explanation:

Step 1:

In the given triangle for angle A, the opposite side has a length of 6 cm, the adjacent side has a length of 8 cm while the hypotenuse of the triangle measures 10 cm. To calculate the tan of angle A we divide the opposite side's length by the adjacent side's length.

$tan A = \frac{oppositeside}{adjacent side}.$

Step 2:

The opposite side's length = 6 cm.

The adjacent side's length = 8 cm.

$tan A = \frac{oppositeside}{adjacent side} = \frac{6}{8} = \frac{3}{4} = 0.75.$

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3 years ago

When you solve using the quadratic formula and you get to

kvasek [131]

Answer:

$\boxed{if \: b= - 2 : {b}^{2} = { (- 2)}^{2} = 4}$

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2 years ago

Part2 of the math homework

Molodets [167]

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3 years ago

Suppose that a box contains 8 cameras and that 4 of them are defective. A sample of 2 cameras is selected at random with replace

Dafna1 [17]

The Expected value of XX is 1.00.

Given that a box contains 8 cameras and that 4 of them are defective and 2 cameras is selected at random with replacement.

The probability distribution of the hypergeometric is as follows:

$P(x,N,n,M)=\frac{\left(\begin{array}{l}M\\ x\end{array}\right)\left(\begin{array}{l}N-M\\ n-x\end{array}\right)}{\left(\begin{array}{l} N\\ n\end{array}\right)}$

Where x is the success in the sample of n trails, N represents the total population, n represents the random sample from the total population and M represents the success in the population.

The probability distribution for X is obtained as below:

From the given information, let X be a random variable, that denotes the number of defective cameras following hypergeometric distribution.

Here, M = 4, n=2 and N=8

The probability distribution of X is obtained below:

The probability distribution of X is,

$P(X=x)=\frac{\left(\begin{array}{l}5\\ x\end{array}\right)\left(\begin{array}{l}8-5\\ 2-x\end{array}\right)}{\left(\begin{array}{l} 8\\ 2\end{array}\right)}$

The probability distribution of X when X=0 is

$\begin{aligned}P(X=0)&=\frac{\left(\begin{array}{l}4\\ 0\end{array}\right)\left(\begin{array}{l}8-4\\ 2-0\end{array}\right)}{\left(\begin{array}{l} 8\\ 2\end{array}\right)}\\ &=\frac{\left(\begin{array}{l}4\\ 0\end{array}\right)\left(\begin{array}{l}4\\ 2\end{array}\right)}{\left(\begin{array}{l} 8\\ 2\end{array}\right)}\\ &=\frac{\left[\left(\frac{4!}{(4-0)!0!}\right)\times \left(\frac{4!}{(4-2)!2!}\right)\right]}{\left(\frac{8!}{(8-2)!2!}\right)}\\ &=0.21\end$

The probability distribution of X when X=1 is

$\begin{aligned}P(X=1)&=\frac{\left(\begin{array}{l}4\\ 1\end{array}\right)\left(\begin{array}{l}8-4\\ 2-1\end{array}\right)}{\left(\begin{array}{l} 8\\ 2\end{array}\right)}\\ &=\frac{\left(\begin{array}{l}4\\ 1\end{array}\right)\left(\begin{array}{l}4\\ 1\end{array}\right)}{\left(\begin{array}{l} 8\\ 2\end{array}\right)}\\ &=\frac{\left[\left(\frac{4!}{(4-1)!1!}\right)\times \left(\frac{4!}{(4-1)!1!}\right)\right]}{\left(\frac{8!}{(8-2)!2!}\right)}\\ &=0.57\end$

The probability distribution of X when X=2 is

$\begin{aligned}P(X=2)&=\frac{\left(\begin{array}{l}4\\ 2\end{array}\right)\left(\begin{array}{l}8-4\\ 2-2\end{array}\right)}{\left(\begin{array}{l} 8\\ 2\end{array}\right)}\\ &=\frac{\left(\begin{array}{l}4\\ 2\end{array}\right)\left(\begin{array}{l}4\\ 0\end{array}\right)}{\left(\begin{array}{l} 8\\ 2\end{array}\right)}\\ &=\frac{\left[\left(\frac{4!}{(4-2)!2!}\right)\times \left(\frac{4!}{(4-0)!0!}\right)\right]}{\left(\frac{8!}{(8-2)!2!}\right)}\\ &=0.21\end$

Use E(X)=∑xP(x) to find the expected values of a random variable X.

The expected values of a random variable X is obtained as shown below:

The expected value of X is,

E(X)=∑xP(x-X)

E(X)=[(0×0.21)+(1×0.57)+(2×0.21)]

E(X)=[0+0.57+0.42]

E(X)=0.99≈1

Hence, the binomial probability distribution of XX when X=0 is 0.21, when X=1 is 0.57 and when X=2 is 0.21 and the expected value of XX is 1.00.

Learn about Binomial probability distribution from here brainly.com/question/10559687

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8 0

2 years ago

What number is 7.9 more than 8.9?

garik1379 [7]

It’s 8.9 because in a simpler way 8 is greater than 7

6 0

2 years ago

Read 2 more answers