Ask question

Ask question

All categories

borishaifa [10]

3 years ago

10

In triangle JKL, the length of side JK is 12 feet, and the length of side KL is 15 feet. Which of the following could be the len

gth of side JL?
A.
27 feet
B.
30 feet
C.
3 feet
D.
4 feet

1 answer:

Andrei [34K]3 years ago

8 0

Answer:

D

Step-by-step explanation:

every 2 sides of a triangle has to be bigger than the 3rd side

You might be interested in

Find the mean of the data in the dot plot below

Firdavs [7]

The mean would be 5
Reason for this is that the mean is when all data is summed up then divided by how many data points there are
So 4+4+7=15
Then divide by 3 would equal 5
I hope this helps!

4 0

2 years ago

Read 2 more answers

What is the difference between d and 7?

Goryan [66]

Answer:

d - 7

Step-by-step explanation:

The word "difference" tells us that we will be using subtraction. Also, subtraction is not the same if you reverse the terms like, x - 2 and 2 - x are not the same. In your question it say "d and 7" so that's the order we use in the algebraic expression.

3 0

1 year ago

Help with 9 and 10 please?

JulsSmile [24]

9. Let x equal the number

Two less than a number is

x - 2

Is more than 15 is

> 15

Put them together for

x-2 > 15

Now solve

x-2 > 15
x > 17

10. Let x equal a number

Seven more than a number is

x+ 7

Is less than or equal to 27 is

<= 27

Put them together for

x + 7 <= 27

Now solve

x+ 7 <= 27
x <= 20

3 0

3 years ago

Find an equation of the line that (a) has the same y-intercept as the line and (b) is parallel to the line .

Zolol [24]

You didn't describe the line. / / / / If the equation of the line is [ y = m x + b ] then [ y=(any number) x+b] has the same y-intercept, and [y=mx+any number] is parallel to it.

8 0

2 years ago

A student S is suspected of cheating on exam, due to evidence E of cheating being present. Suppose that in the case of a cheatin

Kay [80]

Answer:

Step-by-step explanation:

Suppose we think of an alphabet X to be the Event of the evidence.

Also, if Y be the Event of cheating; &

Y' be the Event of not involved in cheating

From the given information:

$P(\dfrac{X}{Y}) = 60\% = 0.6$

$P(\dfrac{X}{Y'}) = 0.01\% = 0.0001$

$P(Y) = 0.01$

Thus, $P(Y') \ will\ be = 1 - P(Y)$

P(Y') = 1 - 0.01

P(Y') = 0.99

The probability of cheating & the evidence is present is = P(YX)

$P(YX) = P(\dfrac{X}{Y}) \ P(Y)$

$P(YX) =0.6 \times 0.01$

$P(YX) =0.006$

The probabilities of not involved in cheating & the evidence are present is:

$P(Y'X) = P(Y') \times P(\dfrac{X}{Y'})$

$P(Y'X) = 0.99 \times 0.0001 \\ \\ P(Y'X) = 0.000099$

(b)

The required probability that the evidence is present is:

P(YX or Y'X) = 0.006 + 0.000099

P(YX or Y'X) = 0.006099

(c)

The required probability that (S) cheat provided the evidence being present is:

Using Bayes Theorem

$P(\dfrac{Y}{X}) = \dfrac{P(YX)}{P(Y)}$

$P(\dfrac{Y}{X}) = \dfrac{P(0.006)}{P(0.006099)}$

$P(\dfrac{Y}{X}) = 0.9838$

5 0

3 years ago