What is the answer to m rqp = 145° and m rqc = 24°. find m cqp?

Why would a bank charge a fee to non-customers to use their ATM?

worty [1.4K]

Answer:

Charging a noncustomer helps pay for the cost of keeping the ATM running and supplied with cash. Customers of the bank already cover the ATM costs with regular fees or by allowing the bank to loan out their money.

Step-by-step explanation:

8 0

3 years ago

Create a polynomial that is added to 5x2 + 6x + 9 to get 7x2 + 9x + 14.

DiKsa [7]

Answer:

Step-by-step explanation:

Subtract first from second

= 2x^2 + 3x + 5

4 0

3 years ago

For one History test, Keith had to answer 40 questions. Of these 40 questions, Keith answered 28 of them correctly. What percent

muminat

Answer:

Keith scored 70% on the History test.

Step-by-step explanation:

We have been given that for History test, Keith had to answer 40 questions. Of these 40 questions, Keith answered 28 of them correctly.

To find the percentage Keith got on History test, we will have to figure out 28 is what percent of 40.

$\text{Percentage scored by Keith on History test}=\frac{28}{40}\times 100$

$\text{Percentage scored by Keith on History test}=0.7\times 100$

$\text{Percentage scored by Keith on History test}=70$

Therefore, Keith scored 70% on the History test.

5 0

3 years ago

Suppose there was a cancer diagnostic test was 95% accurate both on those that do and those do not have the disease. If 0.4% of

Musya8 [376]

Complete Question

Suppose there was a cancer diagnostic test was 95% accurate both on those that do and 90% on those do not have the disease. If 0.4% of the population have cancer, compute the probability that a particular individual has cancer, given that the test indicates he or she has cancer.

Answer:

The probability is $P(C | A) = 0.0042$

Step-by-step explanation:

From the question we are told that

The probability that the test was accurate given that the person has cancer is

$P(A | C) = 0.95$

The probability that the test was accurate given that the person do not have cancer is

$P(A | C') = 0.90$

The probability that a person has cancer is

$P(C) = 0.004$

Generally the probability that a person do not have cancer is

$P(C') = 1- P(C)$

=> $P(C') = 1- 0.004$

=> $P(C') = 0.996$

Generally the probability that a particular individual has cancer, given that the test indicates he or she has cancer is according to Bayes's theorem evaluated as

$P(C | A) = \frac{P(A | C) * P(C)}{P(A|C) * P(C) + P(A| C') * P(C')}$

=> $P(C | A) = \frac{ 0.95 * 0.004 }{ 0.95 * 0.004 + 0.90 * 0.996}$

=> $P(C | A) = 0.0042$

7 0

3 years ago

30 POINTS! Please answer and show work for both!

monitta

The recommended approach for such equations is to eliminate fractions first, by multiplying by the LCM of the denominators. You can simply simplify, then work them as a 2-step linear equation, doing your arithmetic with the fractions as they are.

a) (5x-1)/7 + (4x-3)/2 -(3-2x)/2 = 6 . . . . . given

... 2(5x -1) +7(4x -3) -7(3 -2x) = 84 . . . . . multiply by 14

... 10x -2 +28x -21 -21 +14x = 84 . . . . . . eliminate parentheses

... 52x -44 = 84 . . . . . . . . . . . . . . . . . . . . collect terms

... 52x = 128 . . . . . . . . . . . . . . . . . . . . . . . add 44

... x = 32/13 . . . . . . . . . . . . . . . . . . . . . . . . divide by 52 and simplify

b) (4x−3)/2 − (5−2x)/3 − (3x−4)/3 = 5 . . . given

... 2x -3/2 -5/3 +(2/3)x -x +4/3 = 5 . . . . . . .eliminate parentheses

... (5/3)x -11/6 = 5 . . . . . . . . . . . . . . . . . . . . simplify

... (5/3)x = 41/6 . . . . . . . . . . . . . . . . . . . . . . add 11/6

... x = (41/6)/(5/3) = 41/10 . . . . . . . . . . . . . .divide by the coefficient of x

_____

Using the first method, you're dealing with integers and you can do a lot of the math in your head. Using the second method, you're dealing with fractions and mixed numbers. A calculator can be helpful, especially if it handles arithmetic with fractions nicely, as many graphing calculators do.

6 0

3 years ago