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3 years ago

Use the graph to evaluate the function for the given input value. A)5 B)-5 C)2 D)-2

Mathematics

1 answer:

vampirchik [111]3 years ago

6 0

Answer:

Step-by-step explanation:

I think its 5 correct me if im wrong sorry little bit rsusty at math since last year

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Find the cos of ∠T please

Colt1911 [192]

Answer:

Step-by-step explanation:

Using the cosine ratio

cos T = $\frac{adjacent}{hypotenuse}$ = $\frac{AT}{CT}$ = $\frac{24}{25}$ → A

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3 years ago

How to find AMN and ANM

zysi [14]

The triangle ANM is similar to triangle ACB. That means they share the same angles. Both AMN and ANM are 75.

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4 years ago

Solve the system of equations by the addition method. <br> 8x- 7= -13<br> 5x+ y= 0

kotegsom [21]

Answer:

x=-3/4 y=15/4

Step-by-step explanation:

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3 years ago

For this item, any answers that are not whole numbers should be entered as a decimal, rounded to the tenths place. Brian's schoo

LenaWriter [7]

Answer:

The probability that the locker code begins with a prime number is 40%.
The probability that the last digit of the locker code is an odd number is 40%

Step-by-step explanation:

Prime numbers are the numbers that only divide with 1 or itself. So in this case it is 5 and 7.

The possible combinations are

1*4*3 (with 5) + 1*4*3(with 7) = 24

Total combinations are 5*4*3=60

(24/60)*100 = 40 %

2. Odd numbers are those which don't divide with 2. So in this case it is 6 and 8.

The possible combinations are

1*4*3 (with 6) + 1*4*3(with 8) = 24

Total combinations are 5*4*3=60

(24/60)*100 = 40 %

3 0

3 years ago

A college student is taking two courses. The probability she passes the first course is 0.67. The probability she passes the sec

andrezito [222]

Answer:

0.58 = 58% probability she passes both courses

Step-by-step explanation:

We can solve this question treating the probabilities as a Venn set.

I am going to say that:

Event A: She passes the first course.

Event B: She passes the second course.

The probability she passes the first course is 0.67.

This means that $P(A) = 0.67$

The probability she passes the second course is 0.7.

This means that $P(B) = 0.7$

The probability she passes at least one of the courses is 0.79.

This means that $P(A \cup B) = 0.79$

a. What is the probability she passes both courses

This is $P(A \cap B)$ .

We use the following relation:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

$P(A \cap B) = P(A) + P(B) - P(A \cup B) = 0.67 + 0.7 - 0.79 = 0.58$

0.58 = 58% probability she passes both courses

5 0

3 years ago