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3 years ago

Write an equation in slope inteecept form of the line that passes through the point (-3 , -4) with slope -4

Mathematics

1 answer:

masha68 [24]3 years ago

8 0

Answer:

y=-4x-16

Step-by-step explanation:

explanation above

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Please Simplify the equation (-6) ^-2

xeze [42]

Answer:

1/36

Step-by-step explanation:

(-6) ^-2

The negative in the exponent means in goes from the numerator to the denominator

(-6) ^-2 = 1/ (-6)^2

We know that

(-6)^2 = -6*-6 = 36

(-6) ^-2 = 1/ 36

5 0

3 years ago

Read 2 more answers

Can somebody help with this ?

Igoryamba

You have to calculate the median and which the numbers all

8 0

4 years ago

The American Automobile Association reported that families planning to travel over the Labor Day weekend would spend an average

Anuta_ua [19.1K]

Answer:

0.7486 = 74.86% probability of family expenses for the weekend being more than $580.

Step-by-step explanation:

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 zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

The American Automobile Association reported that families planning to travel over the Labor Day weekend would spend an average of $735.

This means that $\mu = 735$

Assume that the amount spent is normally distributed with a standard deviation of $230.

This means that $\sigma = 230$

What is the probability of family expenses for the weekend being: a. more than $580

This is 1 subtracted by the pvalue of Z when X = 580. So

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

$Z = \frac{580 - 735}{230}$

$Z = -0.67$

$Z = -0.67$ has a pvalue of 0.2514

1 - 0.2514 = 0.7486

0.7486 = 74.86% probability of family expenses for the weekend being more than $580.

5 0

3 years ago

This exercise illustrates that poor quality can affect schedules and costs. A manufacturing process has 120 customer orders to f

Aliun [14]

Answer:

a. P(X = 0) = 0.02586

b. $\mathbf{P(X \leq 2 ) =0.2879}$

c. $\mathbf{P(X \leq 5 ) =0.8387}$

Step-by-step explanation:

From the given information:

a. If the manufacturer stocks 120 components, what is the probability that the 120 orders can be filled without reordering components?

$P(X = 0)=(^{120}_{0}) (0.03)^0 (1-0.03)^{n-0}$

$P(X = 0)=\dfrac{120!}{0!(120-0)!} (0.03)^0 (1-0.03)^{n-0}$

P(X = 0) = 1 × 1 ( 0.97)¹²⁰ ⁻ ⁰

P(X = 0) = 0.02586

b. ) If the manufacturer stocks 122 components, what is the probability that the 120 orders can be filled without reordering components?

$P(X \leq 2 ) = [ P(X=0) + P(X =1) + P(X = 2) ]$

$P(X \leq 2 ) = [(^{122}_{0})(0.03)^0 (0.97)^{122-0}+(^{122}_{1})(0.03)^1 (0.97)^{122-1}+(^{122}_{2})(0.03)^2 (0.97)^{122-2}]$ $P(X \leq 2 ) = [\dfrac{122!}{0!(122-0)! } \times 1 \times (0.97)^{122}+\dfrac{122!}{1!(122-1)! } \times (0.03) (0.97)^{121}+\dfrac{122!}{2!(122-2)! } \times 9 \times 10^{-4} \times (0.97)^{120}]$

$P(X \leq 2 ) = [(1 \times 1 \times 0.02433 )+(122 \times (0.03) \times 0.025083)+(7381 \times 9 \times 10^{-4} \times 0.02586)]$

$\mathbf{P(X \leq 2 ) =0.2879}$

(c) If the manufacturer stocks 125 components, what is the probability that the 120 orders can be filled without reordering components?

$P(X \leq 5 ) = [ P(X=0) + P(X =1) + P(X = 2) +P(X = 3)+P(X = 4)+ P(X = 5) ]$

$P(X \leq 5 ) = [(^{122}_{0})(0.03)^0 (0.97)^{122-0}+(^{122}_{1})(0.03)^1 (0.97)^{122-1}+(^{122}_{2})(0.03)^2 (0.97)^{122-2} + (^{122}_{3})(0.03)^3 (0.97)^{122-3} + (^{122}_{4})(0.03)^4 (0.97)^{122-4}+ (^{122}_{5})(0.03)^5 (0.97)^{122-5}]$ $P(X \leq 5 ) = [\dfrac{122!}{0!(122-0)! } \times 1 \times (0.97)^{122}+\dfrac{122!}{1!(122-1)! } \times (0.03) (0.97)^{121}+\dfrac{122!}{2!(122-2)! } \times 9 \times 10^{-4} \times (0.97)^{120} + \dfrac{122!}{3!(122-3)! }*(0.03)^3(0.97)^{122-3)}+ \dfrac{122!}{4!(122-4)! }*(0.03)^4(0.97)^{122-4)} +\dfrac{122!}{5!(122-5)! } *(0.03)^5(0.97)^{122-5)}]$

$\mathbf{P(X \leq 5 ) =0.8387}$

5 0

3 years ago

Find the perimeter of a square of length of 5cm

Alla [95]

Answer:

20cm

Step-by-step explanation:

If each side of the square = 5 cm, 5 cm times 4 sides = 20 cm.

6 0

4 years ago

Read 2 more answers