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3 years ago

I need help with this

Mathematics

1 answer:

Yanka [14]3 years ago

7 0

An yes

Step-by-step explanation:

2=2 is 5

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Solve for f please :)

kkurt [141]

The answer is A) i think

6 0

3 years ago

The area of a bulletin board is 52 square feet. The length is 3 feet less than4 times the width. Find the length and width

Brilliant_brown [7]

Answer:

The length of the bulletin board is $13\ ft$

The width of the bulletin board is $4\ ft$

Step-by-step explanation:

Let

x-----> the length of the bulletin board

y-----> the width of the bulletin board

we know that

The area of the rectangle (bulletin board) is equal to

$A=xy$

we have

$A=52\ ft^{2}$

$52=xy$ ------> equation A

$x=4y-3$ ------> equation B

substitute equation B in equation A and solve for y

$52=(4y-3)y$

$4y^{2}-3y-52=0$

Using a graphing calculator------> solve the quadratic equation

The solution is

$y=4\ ft$

see the attached figure

Find the value of x

$x=4(4)-3=13\ ft$

6 0

4 years ago

Choose 3 values that would make this inequality true. n - 3 ≤ 10

PSYCHO15rus [73]

Answer:

$n = 12, 11, 10, 9...$

Step-by-step explanation:

Step 1: Solve for n

$n - 3 \leq 10$

$n-3+3\leq 10+3$

$n\leq 13$

Step 2: Choose 3 values

$n\leq 13$

$n = 12, 11, 10, 9...$

Answer: $n = 12, 11, 10, 9...$

5 0

3 years ago

Read 2 more answers

What is the solution to the system of equations shown below?

My name is Ann [436]

Answer:

Step-by-step explanation:

____

Good evening ,

_______________

2(7) + 5(1) + 3(-3) = 10

3(7) - (1) + 4(-3) = 8

5(7) - 2(1) + 7(-3) = 12

Then (7,1,-3) is a solution to system

____

8 0

3 years ago

A shop sells a particular of video recorder. Assuming that the weekly demand for the video recorder is a Poisson variable with t

julia-pushkina [17]

Answer:

a) 0.5768 = 57.68% probability that the shop sells at least 3 in a week.

b) 0.988 = 98.8% probability that the shop sells at most 7 in a week.

c) 0.0104 = 1.04% probability that the shop sells more than 20 in a month.

Step-by-step explanation:

For questions a and b, the Poisson distribution is used, while for question c, the normal approximation is used.

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\lambda}*\lambda^{x}}{(x)!}$

In which

x is the number of successes

e = 2.71828 is the Euler number

$\lambda$ is the mean in the given interval.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

The Poisson distribution can be approximated to the normal with $\mu = \lambda, \sigma = \sqrt{\lambda}$ , if $\lambda>10$ .

Poisson variable with the mean 3

This means that $\lambda= 3$ .

(a) At least 3 in a week.

This is $P(X \geq 3)$ . So

$P(X \geq 3) = 1 - P(X < 3)$

In which:

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

Then

$P(X = x) = \frac{e^{-\lambda}*\lambda^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-3}*3^{0}}{(0)!} = 0.0498$

$P(X = 1) = \frac{e^{-3}*3^{1}}{(1)!} = 0.1494$

$P(X = 2) = \frac{e^{-3}*3^{2}}{(2)!} = 0.2240$

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0498 + 0.1494 + 0.2240 = 0.4232$

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 1 - 0.4232 = 0.5768$

0.5768 = 57.68% probability that the shop sells at least 3 in a week.

(b) At most 7 in a week.

This is:

$P(X \leq 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7)$

In which

$P(X = x) = \frac{e^{-\lambda}*\lambda^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-3}*3^{0}}{(0)!} = 0.0498$

$P(X = 1) = \frac{e^{-3}*3^{1}}{(1)!} = 0.1494$

$P(X = 2) = \frac{e^{-3}*3^{2}}{(2)!} = 0.2240$

$P(X = 3) = \frac{e^{-3}*3^{3}}{(3)!} = 0.2240$

$P(X = 4) = \frac{e^{-3}*3^{4}}{(4)!} = 0.1680$

$P(X = 5) = \frac{e^{-3}*3^{5}}{(5)!} = 0.1008$

$P(X = 6) = \frac{e^{-3}*3^{6}}{(6)!} = 0.0504$

$P(X = 7) = \frac{e^{-3}*3^{7}}{(7)!} = 0.0216$

Then

$P(X \leq 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) = 0.0498 + 0.1494 + 0.2240 + 0.2240 + 0.1680 + 0.1008 + 0.0504 + 0.0216 = 0.988$