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3 years ago

I need help with this problem please help me find the answer

Mathematics

1 answer:

olga55 [171]3 years ago

7 0

It’s 14.

The five numbers will be: 12, 13, 14, 15, and 17 so 14 is the median.

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) At a certain college, students are allowed to choose between online and in-person learning. 70% of the students choose online

lubasha [3.4K]

Answer:

Probability = 0.119

Step-by-step explanation:

P (Coronavirus) = P(Person online & corona) or P(Person offline & corona)

(0.70 x 0.02) + (0.30 x 0.35)

0.014 + 0.105

0.119

4 0

2 years ago

A company earns a profit of $100 the first month it is in business. Every month after that, the company earns a profit that is 1

Inga [223]

11/2 of $100 is $550
($550 x 4) + $100 = <span>$2,300
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3 years ago

Painting a Room

aniked [119]

Answer:

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Step-by-step explanation:

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2 years ago

Read 2 more answers

Use the table to write and solve an equation to find the number of

exis [7]

Answer:

$c = 11$

Step-by-step explanation:

Given

$Cars\ Sold:-: 8; 9; 12; c$

Required

What value of c makes the mean = 10

Mean is calculated as

$Mean = \frac{\sum x}{n}$

In this case;

Mean = 10 and n = 4;

So, we have:

$10 = \frac{8+ 9 + 12 + c}{4}$

$10 = \frac{29+ c}{4}$

Multiply both sides by 4

$10 * 4 = 29 + c$

$40 = 29 + c$

Solve for c

$c = 40 - 29$

$c = 11$

4 0

3 years ago

A blueberry pie is shot out of a pie cannon from a height of 6m and with a velocity of 60 m/s. How many seconds will pass before

sveta [45]

Answer: 1.106 s

Step-by-step explanation:

This situation is related to projectile motion and one the equation that models the height of the blueberry pie in time is:

$y=y_{o}+V_{o}sin(\theta) t-\frac{1}{2}gt^{2}$

Where:

$y=0 m$ is the blueberry pie final height (when it hits the ground)

$y_{o}=6 m$ is the blueberry pie initial height

$V_{o}=60 m/s$ is the blueberry pie initial velocity

$\theta=0 \°$ is the angle, assuming the pie was shot horizontally

$t$ is the time

$g=9.8 m/s^{2}$ is the acceleration due gravity

Rewriting the equation:

$0=y_{o}-\frac{1}{2}gt^{2}$

Isolating $t$ :

$t=\sqrt{\frac{2y_{o}}{g}}$

$t=\sqrt{\frac{2(6 m)}{9.8 m/s^{2}}}$

Finally:

$t=1.106 s$

5 0

3 years ago