HELPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP MEEE PLSSSS

Mathematics

1 answer:

allochka39001 [22]3 years ago

4 0

Answer:

Ur question? Itz incomplete

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76 is 95% of what number and my second question is what percent of 17 is 20.4

Oksi-84 [34.3K]

Firstly. make your unknown number X.

95% can be represented as 95/100 or 0.95

so 95% of X is 76...

0.95 × X = 0.76
X = 0.76 ÷ 0.95
X = 0.8

_______________________________________________

for the second question.

(17 ÷ 20.4) and times it by 100 to get a percentage.

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3 years ago

Please help - 30 points.<br> I will mark the best answer Brainliest.

Aloiza [94]

Answer: the first part of this question is C, the second part of this question is D

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4 years ago

What is -6 -4(3×2) + 5 ²

coldgirl [10]

Remember PEMDAS: Order of operations.
1- parentheses
2- exponents
3- multiplication
4- division
5- addition
6- subtraction

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3 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:

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