Ask question

Ask question

All categories

3 years ago

7

Select the values that make the inequality -7w>-63 true then write an equivalent inequality, in terms of w

1 answer:

AlexFokin [52]3 years ago

5 0

put these values in place of w in the inequality u will get Verified.

Step-by-step explanation:

5>4>3>2>1>0>-1>-2>-3.........>-63

here the numbers which come before - 63

are greater than it.

Hope u understood.

You might be interested in

Please help me!!!!!!!!!!

Kipish [7]

Answer:

80 is the kilometers question

8 0

2 years ago

The quartiles of any distribution are the values with cumulative proportions 0.25 and 0.75. What are the quartiles of the standa

choli [55]

$\mathbb P(X\le Q_1)=F_X(Q_1)=0.25\implies Q_1={F_X}^{-1}(0.25)\approx-0.6745$

where $F_X(x)$ is the CDF for $X\sim\mathcal N(0,1)$ .

By symmetry of the distribution, you then have

$Q_3=-Q_1\approx0.6745$

6 0

3 years ago

How many times does 10 go into 100,000,000,000,000,000?

PilotLPTM [1.2K]

100,000,000. this is how many times 1<span>0 goes into 100,000,000,000,000,000</span>

4 0

2 years ago

Read 2 more answers

Required information Skip to question A die (six faces) has the number 1 painted on two of its faces, the number 2 painted on th

grigory [225]

Answer:

The change to the face 3 affects the value of P(Odd Number)

Step-by-step explanation:

Analysing the question one statement at a time.

Before the face with 3 is loaded to be twice likely to come up.

The sample space is:

$S = \{1,1,2,2,2,3\}$

And the probability of each is:

$P(1) = \frac{n(1)}{n(s)}$

$P(1) = \frac{2}{6}$

$P(1) = \frac{1}{3}$

$P(2) = \frac{n(2)}{n(s)}$

$P(2) = \frac{3}{6}$

$P(2) = \frac{1}{2}$

$P(3) = \frac{n(3)}{n(s)}$

$P(3) = \frac{1}{6}$

P(Odd Number) is then calculated as:

$P(Odd\ Number) = P(1) + P(3)$

$P(Odd\ Number) = \frac{1}{3} + \frac{1}{6}$

Take LCM

$P(Odd\ Number) = \frac{2+1}{6}$

$P(Odd\ Number) = \frac{3}{6}$

$P(Odd\ Number) = \frac{1}{2}$

After the face with 3 is loaded to be twice likely to come up.

The sample space becomes:

$S = \{1,1,2,2,2,3,3\}$

The probability of each is:

$P(1) = \frac{n(1)}{n(s)}$

$P(1) = \frac{2}{7}$

$P(2) = \frac{n(2)}{n(s)}$

$P(2) = \frac{3}{7}$

$P(3) = \frac{n(3)}{n(s)}$

$P(3) = \frac{1}{7}$

$P(Odd\ Number) = P(1) + P(3)$

$P(Odd\ Number) = \frac{2}{7} + \frac{1}{7}$

Take LCM

$P(Odd\ Number) = \frac{2+1}{7}$

$P(Odd\ Number) = \frac{3}{7}$

Comparing P(Odd Number) before and after

$P(Odd\ Number) = \frac{1}{2}$ --- Before

$P(Odd\ Number) = \frac{3}{7}$ --- After

<em>We can conclude that the change to the face 3 affects the value of P(Odd Number)</em>

7 0

3 years ago

Howard has a garden in the shape of a rectangle. The length is 5.4 meters and the width is 1.5 meters. Howard will increase both

Alla [95]

Please find the attachment below for the solution...

4 0

3 years ago