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2 years ago

8

There are five envelopes on a table. Four of the envelopes are marked 8, 2, 4 and 1. You are told that the mean of the numbers o

n the envelopes is 4. What is the number on the fifth envelope?

1 answer:

Lostsunrise [7]2 years ago

3 0

Answer: 16

Step-by-step explanation:

You might be interested in

Which of the following statements best describes the location of -(-1) on a number line?

Novosadov [1.4K]

Answer:

Step-by-step explanation:

-(-1) is 1 because if you multiply by a negative number becomes positive

so

it is 1 unit to the right of 0

8 0

3 years ago

4(x - 2) = -12, x = 4

rewona [7]

Answer:

x= -1

-1 =4

Step-by-step explanation:

3 0

3 years ago

Pizza by the Pan sold 4 dozen pizzas in the afternoon. That night, they sold 2.5 times as many pizzas as they did during the aft

Sav [38]

Answer:

they sold 14 pizzas all day.

Step-by-step explanation:

4 x 2 = 8

0.5 means half of what they sold in the afternoon

4 ÷ 2 = 2

now add them both together to get the number of pizzas they sold at night.

8 + 2 = 10

now add the afternoon and night together

10 + 4 = 14

answer: 14 pizzas

6 0

3 years ago

Rewrite the following equation in slope-intercept form.

Juli2301 [7.4K]

Answer:

y=-18/5(x)-7/5

Step-by-step explanation:

7 0

3 years ago

A rectangular lamina of uniform density is situated with opposite corners at (0,0) and (15,4). calculate its radii of gyration a

Sphinxa [80]

First, you have to find the moment of inertia along the x and y axes. Constant density is denoted as k.

$I_{x}= \int\limits^15_0\int\limits^4_0 {k y^{2} } \, dx dy= \frac{1}{3}k (15-4)^4=4880.33k$

$I_{y}= \int\limits^15_0\int\limits^4_0 {k x^{2} } \, dx dy= \frac{1}{3}k (15-4)^4=4880.33k$

Then, the radii of gyration for

x = √[I_x/m]
y = [I_y/m]

where m = k(15-4)² = 121k. Then,

x = y = [4880.33k/121k] = 40.33

I hope I was able to help you. Have a good day.

7 0

3 years ago