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3 years ago

I don’t get none of the work it is very hard to do

Mathematics

2 answers:

Misha Larkins [42]3 years ago

7 0

Answer:

1÷2×12

pls mark as brainliest

Snowcat [4.5K]3 years ago

3 0

Answer:

Step-by-step explanation:

So first you would do what it stays on the screen

Then you would have 1/2*12

Now 1/2 of 12 is 6 because 6+6=12

So therefore your blank box is the number 6

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What is the value of the expression? −3/4+(1/10÷2/5) Enter your answer, as a fraction in simplest form, in the box.

blagie [28]

-11/20 would be the value of the expression

8 0

3 years ago

Read 2 more answers

Solve for m: 8(m + 2) = 3(12 - 4m)

Tema [17]

Answer:

m=1

Step-by-step explanation:

step 1: 8(m + 2) = 3(12 - 4m) remove the parentheses

step 2: 8 m + 2 = 3 12 - 4m move the terms

step 3: 8m+ 12m= 36 - 16 collect like terms and calulate

step 4: 20m=20 divide both sides

solution: m=1

3 0

3 years ago

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A professor pays 25 cents for each blackboard error made in lecture to the student who pointsout the error. In a career ofnyears

marta [7]

Answer:

(a) The probability that Y₂₀ exceeds 1000 is 3.91 × 10⁻⁶.

(b) n = 28.09

Step-by-step explanation:

The random variable Yₙ is defined as the total numbers of dollars paid in n years.

It is provided that Yₙ can be approximated by a Gaussian distribution, also known as Normal distribution.

The mean and standard deviation of Yₙ are:

$\mu_{Y_{n}}=40n\\\sigma_{Y_{n}}=\sqrt{100n}$

(a)

For n = 20 the mean and standard deviation of Y₂₀ are:

$\mu_{Y_{n}}=40n=40\times20=800\\\sigma_{Y_{n}}=\sqrt{100n}=\sqrt{100\times20}=44.72\\$

Compute the probability that Y₂₀ exceeds 1000 as follows:

$P(Y_{n}>1000)=P(\frac{Y_{n}-\mu_{Y_{n}}}{\sigma_{Y_{n}}}>\frac{1000-800}{44.72})\\=P(Z> 4.47)\\=1-P(Z$

**Use a z table for probability.

Thus, the probability that Y₂₀ exceeds 1000 is 3.91 × 10⁻⁶.

(b)

It is provided that P (Yₙ > 1000) > 0.99.

$P(Y_{n}>1000)=0.99\\1-P(Y_{n}$

The value of z for which P (Z < z) = 0.01 is 2.33.

Compute the value of n as follows:

$z=\frac{Y_{n}-\mu_{Y_{n}}}{\sigma_{Y_{n}}}\\2.33=\frac{1000-40n}{\sqrt{100n}}\\2.33=\frac{100}{\sqrt{n}}-4\sqrt{n} \\2.33=\frac{100-4n}{\sqrt{n}} \\5.4289=\frac{(100-4n)^{2}}{n}\\5.4289=\frac{10000+16n^{2}-800n}{n}\\5.4289n=10000+16n^{2}-800n\\16n^{2}-805.4289n+10000=0$

The last equation is a quadratic equation.

The roots of a quadratic equation are:

$n=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

a = 16

b = -805.4289

c = 10000

On solving the last equation the value of n = 28.09.

8 0

3 years ago

Which inequality is shown?

Nimfa-mama [501]

Answer:

It should be J

Step-by-step explanation:

The line is solid indicating it is a greater than/less than/equal relationship and the shaded area indicated a positive y (so greater than)

5 0

3 years ago

WILL MARK BRAINLIEST what is the standard deviation for the normal distribution shown at the right?

MrRissso [65]

Answer: 916. Because if you do 856+60=916.

Step-by-step explanation: The standard normal distribution is a normal distribution with a mean of zero and standard deviation of 1. Formula: and natural statistics x and x2. The dual expectation parameters for normal distribution are η1 = μ and η2 = μ2 + σ2.

8 0

3 years ago