All categories

3 years ago

Expand and simplify (X+5) (x+7)

Mathematics

1 answer:

Hoochie [10]3 years ago

4 0

Answer:

x^2+12x+35

Step-by-step explanation:

x(x+7)+5(x+7)

x^2+7x+5x +35

x^2+12x+35

You might be interested in

What is the value of x in the equation below?

11111nata11111 [884]

Answer: B 9.25

Step-by-step explanation:

5 0

3 years ago

Find the value of x.

padilas [110]

To find your missing angle:
180-126 = 54
so 9x must equal 54
54= 9x
6 = x

3 0

3 years ago

What is the formula for mortgages?

professor190 [17]

Answer:

Loan payment = Loan amount / Discount factor

Number of Periodic Payments (n) = Payments per year times number of years. Periodic Interest Rate (i) = Annual rate divided by number of payments per. Discount Factor (D) = {[(1 + i) ^n] - 1} / [i(1 + i)^n]

Step-by-step explanation:

7 0

3 years ago

1. A report from the Secretary of Health and Human Services stated that 70% of single-vehicle traffic fatalities that occur at n

Nuetrik [128]

Using the binomial distribution, it is found that there is a 0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.

For each fatality, there are only two possible outcomes, either it involved an intoxicated driver, or it did not. The probability of a fatality involving an intoxicated driver is independent of any other fatality, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

x is the number of successes.
n is the number of trials.
p is the probability of a success on a single trial.

In this problem:

70% of fatalities involve an intoxicated driver, hence $p = 0.7$ .

A sample of 15 fatalities is taken, hence $n = 15$ .

The probability is:

$P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)$

Hence

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 10) = C_{15,10}.(0.7)^{10}.(0.3)^{5} = 0.2061$

$P(X = 11) = C_{15,11}.(0.7)^{11}.(0.3)^{4} = 0.2186$

$P(X = 12) = C_{15,12}.(0.7)^{12}.(0.3)^{3} = 0.1700$

$P(X = 13) = C_{15,13}.(0.7)^{13}.(0.3)^{2} = 0.0916$

$P(X = 14) = C_{15,14}.(0.7)^{14}.(0.3)^{1} = 0.0305$

$P(X = 15) = C_{15,15}.(0.7)^{15}.(0.3)^{0} = 0.0047$

Then:

$P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15) = 0.2061 + 0.2186 + 0.1700 + 0.0916 + 0.0305 + 0.0047 = 0.7215$

0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.

A similar problem is given at brainly.com/question/24863377

5 0

3 years ago

3. Convert 0.002 kilojoules to decijoules:

k0ka [10]

Answer:

20 is the answer.

5 0

3 years ago

Read 2 more answers