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2 years ago

During a period of 30 working days, Deborah was late 80% of the time. How many days was she late?

1 answer:

8 0

Fairly straightforward question. She works 30 days and is late 80% or 0.8
We can model our equation
Num days late = late percentage * num working days
Basically fine portion of working days that she was late.

Num days late = 0.8 * 30=24

So Deborah was late 24 out of 30 working days. Would appreciate a brainliest

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Can someone please help me.. very quickly .. thank you

Alisiya [41]

X/8 + 12 = 16
start by subtracting twelve from both sides
x/8 + 12 = 16
- 12 -12
You're left with x/8 = 4
Multiply both sides by 8
8× x/8 = 4 ×8
Your answer is x = 32
If you need to simplify, the answer is 2.

3 0

3 years ago

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The manager at the local auto shop has found that the probability that a car brought into the shop requires an oil change is 0.8

Ber [7]

Answer:

0.85

Step-by-step explanation:

Given two events A and B, the probability that either A or B occurs is given by:

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

where

$p(A)$ the probability that A occurs

$p(B)$ is the probability that B occurs

$p(A\cap B)$ is the probability that both A and B occur at the same time

In this problem, we know the following facts:

$p(o) = 0.83$ is the probability that the car requires an oil change

$p(b)=0.17$ is the probability that the car requires a brake repair

$p(o\cap b) = 0.15$ is the probability that the car requires both an oil change and brake repair

Therefore, the probability that either o (car requiring oil change) or b (car requiring brake repait) occur is:

$p(o\cup b)=p(o)+p(b)-p(o\cap b)=0.83+0.17-0.15 = 0.85$

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2 years ago

Determine the value of A <br> Solve it please

Vitek1552 [10]

A= 30
Explanation: 88-58=30

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2 years ago

Line AB contains points A(4, 5) and B(9, 7). What is the slope of AB?

horsena [70]

The answer is 2/5 which is C. You're welcome

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3 years ago

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What is the arc measure of BDC in degrees?<br> (4k + 159)<br> P<br> (2k + 153)

andre [41]

<h2>Explanation:</h2><h2></h2>

The diagram is missing but I'll assume that the arc BDC is:

$(4k + 159)^{\circ}$

And another arc, let's call it FGH. measures:

$(2k + 153)^{\circ}$

If those arc are equal, then this equation is true:

$(4k + 159)^{\circ}=(2k + 153)^{\circ} \\ \\ (4k + 159)=(2k + 153) \\ \\ \\ Solving \ for \ k: \\ \\ 4k-2k=153-159 \\ \\ 2k=-6 \\ \\ k=-\frac{6}{2} \\ \\ k=-3$

Substituting k into the first equation:

$\angle BDC=(4(-3)+159)^{\circ} \\ \\ \angle BDC=(-12+159)^{\circ} \\ \\ \boxed{\angle BDC=147^{\circ}}$

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2 years ago

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