Solve y∕2.5 = 5 for y.

*EASY 7TH GRADE MATH*

meriva

Answer:

A rectangle that has the same volume can be the opposite of that.

Step-by-step explanation:

The length = 4

the width = 20

the height = 6

Hope this helps,

v= l ×w×h

ヾ(・ω・`)ノ

8 0

3 years ago

Two airplanes leave an airport at the same time and travel in opposite directions. One plane travels 66 kmh faster than the othe

goldfiish [28.3K]

Answer:

The rate of planes: 926 km/h and 860 km/h

=========================

<h3>Given</h3>

Two planes travel in opposite directions,
One plane travels 66 km/h faster than the other,
The distance between the two planes after 6 hours is 10716 km.

The rate of of each plane

<h3>Solution</h3>

The distance traveled by two planes in an hour is:

10716/6 = 1786 km/h

Let the rate of one plane be x, then the other plane's rate is x - 66.

The sum of two rates is the distance traveled in one hour:

x + x - 66 = 1786
2x - 66 = 1786
2x = 1786 + 66
2x = 1852
x = 1852/2
x = 926

So the rate of one plane is 926 km/h and of the other plane is:

926 - 66 = 860 km/h

3 0

2 years ago

As a company manager for claimstat corporation there is a 0.40 probability that you will be promoted this year. there is a 0.72

Charra [1.4K]

Suppose $P$ denotes the event of getting promoted and $R$ the event of getting a raise. Then

$\mathbb P(P\cup R)=\mathbb P(P)+\mathbb P(R)-\mathbb P(P\cap R)$
$\implies0.72=0.40+\mathbb P(R)-0.25$
$\implies\mathbb P(R)=0.57$

Now it sounds like you're asked to find $\mathbb P(R\mid P)$ , which the definition of conditional probability says is equivalent to

$\mathbb P(R\mid P)=\dfrac{\mathbb P(R\cap P)}{\mathbb P(P)}=\dfrac{0.25}{0.40}=0.625$

3 0

3 years ago

Below are data on 18 people who fell ill from an incident of food poisoning. The data give the incubation period (the time in ho

Sveta_85 [38]

Answer:

$\bar X= \frac{15+16+18+19+20+20+21+28+32+34+36+43+46+46+48+48+72+88}{18}= 36.11$

$Min =15$

$Q_1 = \frac{20+20}{2}=20$

$Median= Q_2= \frac{32+34}{2}=33$

$Q_3 = \frac{46+46}{2}=46$

$Max= 88$

The IQR is $IQR= Q_3 -Q_1= 46-20=26$

We can find the usual limits for the values and we got:

$Lower = Q_1 -1.5 IQR = 20 -1.5*26=-19$

$Upper = Q_3 +1.5 IQR = 46 +1.5*26=85$

So then the potential outliers for this case is just 88>85

Step-by-step explanation:

For this case we have the following data:

15 16 18 19 20 20 21 28 32 34 36 43 46 46 48 48 72 88

The sample size is n =18

We can calculate the mean with the following formula:

$\bar X= \frac{\sum_{i=1}^n X_i}{n}$

And if we replace we got:

$\bar X= \frac{15+16+18+19+20+20+21+28+32+34+36+43+46+46+48+48+72+88}{18}= 36.11$

The minimum for this case is $Min =15$

Now we can find the 5 number summary.

For the first quartile we work with the first 10 observations: 15 16 18 19 20 20 21 28 32 34. And Q1 would be the average between the 5th and 6th position of the data ordered, on this case:

$Q_1 = \frac{20+20}{2}=20$

For the median since we have an even number for the sample size would be the average between the 9th and the 10th position from the dataset ordered and we got:

$Median= Q_2= \frac{32+34}{2}=33$

For the third quartile we work with the last 10 observations: 32 34 36 43 46 46 48 48 72 88. And Q3 would be the average between the 5th and 6th position of the data ordered, on this case:

$Q_3 = \frac{46+46}{2}=46$

The maximum is $Max= 88$

The IQR is $IQR= Q_3 -Q_1= 46-20=26$

We can find the usual limits for the values and we got:

$Lower = Q_1 -1.5 IQR = 20 -1.5*26=-19$

$Upper = Q_3 +1.5 IQR = 46 +1.5*26=85$

So then the potential outliers for this case is just 88>85

3 0

3 years ago

What are three fractions that are equivalent to 1.

tankabanditka [31]

2/2 , 3/3 , 4/4 ( anything over its self)

5 0

3 years ago