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3 years ago

Could someone please help and explain?

Mathematics

1 answer:

NeTakaya3 years ago

8 0

Answer:

Step-by-step explanation:

Basically, a number which is divisible by 4 formed by the last two digits in a number is divisible by 4. So, we can see 28 is divisible by 4... Therefore the anser is simply "2".

328÷4=82. :)) Hope it helps!

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The cost of 12 oranges and 6 apples is $5.38 Eight oranges and 5 apples cost $3.68 Find the cost of each

Ksju [112]

Let the oranges cost = x

Let the apples cost = y

According to given conditions, equation becomes:

$12x+6y=5.38$ .... (1)

$8x+5y=3.68$ .....(2)

Multiplying equation (1) by 8 and equation (2) by 12 to get y

96x+48y=43.04 .... (3)

96x+60y=44.16 .... (4)

Now subtracting both equations we get,

12y = 1.12

y = 0.093

Hence, cost of 1 apple = $0.093

As 12x+6y=5.38

12x+6(0.093)=5.38

12x=4.822

x = 0.40

Hence, cost of 1 orange = $0.40

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4 years ago

Mr. and Mrs. Gomez are buying carpet for their square-shaped bedroom, which measures 8 yd along each wall. The carpet is $ 22

Paha777 [63]

Answer:

$\$1,664$

Step-by-step explanation:

step 1

Find the area of the square-shaped bedroom

$A=8^{2}=64\ yd^{2}$

step 2

calculate the total spend

Multiply the area of the bedroom by $(22+4)=26\frac{\$}{yd^{2} }$

$64*26=\$1,664$

3 0

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:

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