All categories

3 years ago

HELP PLEASE I NEED HELP

Mathematics

2 answers:

polet [3.4K]3 years ago

8 0

47 and 4 are the right answers

andriy [413]3 years ago

6 0

The answer to this question using PEMDAS is

47.4

You might be interested in

A credit card company applies a 6% fee on all transactions under $10. You purchased coffee for $4.80. What fee amount will be ap

Romashka [77]

Answer:

$5.08

Step-by-step explanation:

Step one:

given data

We are told that a credit card company applies a 6% fee on all transactions under $10

since the cost of coffee bellow $10 that is $4.80, the 6% fee is applicable

=6% of 4.80

=6/100*4.80

=0.06*4.8

=$0.288

Step two:

the total fee will be

=0.288+4.80

= $5.08

7 0

3 years ago

The expression below is a sum of cubes. 125x^3 + 144 A. True B. False

pishuonlain [190]

It would be false bc i just took it and it was right

8 0

3 years ago

Read 2 more answers

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

What is the value of x?

user100 [1]

Answer:

1/2

Step-by-step explanation:

isolate the variable by dividing each side by factors that don't contain the variable

7 0

3 years ago

Read 2 more answers

Write the equation of the line that passes through the points (- 8, - 6) and (- 8, 7) . Put your answer in fully reduced point-s

Darya [45]

Answer:

Equation of the given straight line is x = -8

Step-by-step explanation:

By the given information we have,

The line is vertical that means it's slope is not defined therefore the equation of the line is given by, x = c where c is the x coordinate.

Therefore x = -8.

4 0

3 years ago