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3 years ago

Evaluate the value of a÷ ( b+c) when a= 51 , b= 15 & c= 2 .

Mathematics

1 answer:

mel-nik [20]3 years ago

4 0

Answer:

by substituting the values we get

51÷(15+2)

51÷17=3

hope it helps you mate

please mark me as brainliast

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Sheree smith purchased a new SUV for $32,984. He received a $49.95 rebate from the manufacturer and a $38.95 rebate from the sto

Ad libitum [116K]

Answer:

33,072.9

Step-by-step explanation:

32,984+49.95+38.95

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4 years ago

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3 years ago

Read 2 more answers

Which equation represents the line that passes through the point (4, -5) and is perpendicular to the line x + 2y = 5?

Lerok [7]

Answer:

y = 2x - 13

Step-by-step explanation:

Equation of a line is y = mx + c, m is the gradient and c is the intercept

The line passes through points 4 and -5, x is 4 and y is -5

-5 = 4m + c

When two lines are perpendicular, the products of their gradients are equal to -1, m1 * m2 = -1

x + 2y = 5

2y = -x + 5

y = (-1/2 * x) + 5

therefore m = -1/2

m1 * m2 = -1

m * -1/2 = -1

-m = -2 , therefore m = 2

-5 = 4 * 2 + c

c = -5 - 8, which is -13

Therefore the equation for the line is

y = 2x - 13

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3 years ago

Quadric formula to solve 9x^2+6x-17=0

raketka [301]

Answer:

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Step-by-step explanation:

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3 years ago

Some parts of California are particularly earthquake- prone. Suppose that in one metropolitan area, 25% of all homeowners are in

aleksandrvk [35]

Answer:

a. Binomial random variable (n=4, p=0.25)

b. Attached.

c. X=1

Step-by-step explanation:

This can be modeled as a binomial random variable, with parameters n=4 (size of the sample) and p=0.25 (proportion of homeowners that are insured against earthquake damage).

a. The probability that X=k homeowners, from the sample of 4, have eartquake insurance is:

$P(x=k) = \dbinom{n}{k} p^{k}(1-p)^{n-k}\\\\\\P(x=k) = \dbinom{4}{k} 0.25^{0}\cdot0.75^{4}$

The sample space for X is {0,1,2,3,4}

The associated probabilties are:

$P(x=0) = \dbinom{4}{0} p^{0}(1-p)^{4}=1*1*0.3164=0.3164\\\\\\P(x=1) = \dbinom{4}{1} p^{1}(1-p)^{3}=4*0.25*0.4219=0.4219\\\\\\P(x=2) = \dbinom{4}{2} p^{2}(1-p)^{2}=6*0.0625*0.5625=0.2109\\\\\\P(x=3) = \dbinom{4}{3} p^{3}(1-p)^{1}=4*0.0156*0.75=0.0469\\\\\\P(x=4) = \dbinom{4}{4} p^{4}(1-p)^{0}=1*0.0039*1=0.0039\\\\\\$

b. The histogram is attached.

c. The most likely value for X is the expected value for X (E(X)).

Is calculated as:

$E(X)=np=4\cdot0.25=1$

6 0

3 years ago