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3 years ago

Amarosa Logistics purchased a new pick-up for the office employees to use for $28,000.00. The vehicle is expected to last for

Mathematics

1 answer:

Lena [83]3 years ago

5 0

Answer:

I think B but I am not sure.

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What is the measure of x? A. 144° B. 72° C. 36° D. 18°

dolphi86 [110]

Here, 4x + x =180 [ Supplementary angles ]
5x = 180
x = 180/5
x = 36

In short, Your Answer would be Option C

Hope this helps!

4 0

4 years ago

Read 2 more answers

Steve can complete the 100m dash in 10 seconds while Paul can run it in 12 seconds. How does Steve's time compare to Paul's?

Lana71 [14]

Let

x---------> Steve's time

y--------> Paul's time

we know that

$x=10\ s\\y=12\ s$

$x < y$ ------> Steve is faster than Paul

Divide x by y

$\frac{10}{12}=\frac{5}{6}$

therefore

the answer is the option

A. Steve is 5⁄6 as fast as Paul

8 0

3 years ago

Read 2 more answers

Can someone please help me with this???

8_murik_8 [283]

there would be 38 woman

8 0

4 years ago

Cystic fibrosis (CF) is a hereditary lung disorder that often results in death. It can be inherited only if both parents are car

OLEGan [10]

Answer:

0.9959 = 99.59% probability that a randomly chosen person of European ancestry does not carry an abnormal CF gene given that he /she tested negative.

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Tested negative

Event B: Does not carry an abdornal CF gene.

Probability of a negative test:

10% of 1/25 = 0.04 = 4%

100% of 100 - 4 = 96%. So

$P(A) = 0.1*0.04 + 1*0.96 = 0.964$

Negative test and not carrying an abdornal gene:

100% of 96%. So

$P(A \cap B) = 1*0.96 = 0.96$

Compute the probability that a randomly chosen person of European ancestry does not carry an abnormal CF gene given that he /she tested negative.

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.96}{0.964} = 0.9959$

0.9959 = 99.59% probability that a randomly chosen person of European ancestry does not carry an abnormal CF gene given that he /she tested negative.

8 0

3 years ago

Helllpppp!! Plz! This is 6th grade work!

Zarrin [17]

In the question it gives you what "s" is which is 3, so plug that in instead of s.

c = 25s²
c = 25(3²)
c = 25(9)
c = 225 << your answer

hope this helps, God bless!

6 0

3 years ago