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3 years ago

What is the simplified expression for 8x + 6 - 14 + 7x - 3y

Mathematics

1 answer:

alexira [117]3 years ago

6 0

Simplify the expression.

15x − 3y − 8

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A set of 3 pens costs $1.68. At that unit price, how much would one pen cost?

Tanzania [10]

Answer:

$.56

Step-by-step explanation:

To find the cost of one pen, take the total cost and divide by the number of pens

$1.68/ 3

$.56 per pen

8 0

1 year ago

An amphitheatre has 24 seats in the first row, 28 in the second, 32 in the third, and so on, for 29 rows. How many seats are

melomori [17]

The answer is 2,320 seats in the amphitheater. Since the seats are increasing at a rate of 4 more seats each row, then just add 4 to the number for each past row starting at 32 since that's where the rest is unknown. Once you find all he numbers just add them up to get your answer.
I hope I helped!

3 0

2 years ago

Plz answer ASAP<br>what is the distance between the lions and monkeys?

3241004551 [841]

im pretty sure you got it right with 40!!

5 0

2 years ago

In a certain city, the daily consumption of electric power, in millions of kilowatt-hours, is a random

kupik [55]

Answer:

a) $\alpha = 3, \beta = 2$

b) 0.0620

Step-by-step explanation:

We are given the following in the question:

Population mean, $\mu$ = 6

Variance, $\sigma^2$ = 12

a) Value of $\alpha, \beta$

We know that

$\alpha \beta = \mu = 6\\\alpha \beta^2 = \sigma^2 = 12$

Dividing the two equations, we get,

$\dfrac{\alpha\beta^2}{\alpha\beta} = \dfrac{12}{6}\\\\\Rightarrow \beta = 2\\\alpha \beta = 6\\\Rightarrow \alpha = 3$

b) probability that on any given day the daily power consumption will exceed 12 million kilowatt hours.

We can write the probability density function as:

$f(x,3,2) = \dfrac{1}{2^{3}(3-1)!}x^{3-1}e^{-\frac{x}{2}}, x > 0\\\\f(x,3,2) = \dfrac{1}{16}x^{2}e^{-\frac{x}{2}}, x > 0$

We have to evaluate:

$P(x >12)\\\\= \dfrac{1}{16}\displaystyle\int^{\infty}_{12}f(x)dx\\\\=\dfrac{1}{16}\bigg[-2x^2e^{-\frac{x}{2}}-2\displaystyle\int xe^{-\frac{x}{2}}dx}\bigg]^{\infty}_{12}\\\\=\dfrac{1}{8}\bigg[x^2e^{-\frac{x}{2}}+4xe^{-\frac{x}{2}}+8e^{-\frac{x}{2}}\bigg]^{\infty}_{12}\\\\=\dfrac{1}{8}\bigg[(\infty)^2e^{-\frac{\infty}{2}}+4(\infty)e^{-\frac{\infty}{2}}+8e^{-\frac{\infty}{2}} -( (12)^2e^{-\frac{12}{2}}+4(12)e^{-\frac{12}{2}}+8e^{-\frac{12}{2}})\bigg]\\\\=0.0620$

0.0620 is the required probability that on any given day the daily power consumption will exceed 12 million kilowatt hours.

3 0

3 years ago

The answers to 19 & 20

hodyreva [135]

Ighty thanks ;;;;;;))))))

6 0

3 years ago