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3 years ago

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The data represents the time, in minutes, spent reading a political blog in a day. Construct a frequency distribution using 5 cl

asses. In the table, include the midpoints, relative frequencies, and cumulative frequencies. Which class has the greatest frequencies and which has the least frequency?

1 answer:

Vinil7 [7]3 years ago

6 0

Answer:

Step-by-step explanation:

7

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Each side of a square is lengthened by 5 inches. the area of this new, larger square is 36 square inches. find the length of a

butalik [34]

The formula of the area is:

$A = s^{2}$

Where s stands for side.

So if the area is 36, the lenght of the side must be:

$s = \sqrt{36}$

$\bf~s = 6$

So if 6 is the side of the original square plus 5, than the original lenght of the side is:

6 - 5 = 1 inch

Hope it helped,

BioTeacher101

6 0

4 years ago

How is this problem done ? <br><br> 9+2|4x−3|=19

grandymaker [24]

Add 9+2 when you have the answer times it by 4 then minus 3 , wow I'm so confused i'm in middle school ha thought i could do it

6 0

3 years ago

Earth rotates on an axis through its poles. The distance from the axis to a location on Earth at 40 degrees north latitude is ab

murzikaleks [220]

Answer: v = 793.77 mph

Step-by-step explanation: Linear speed of a circular movement is calculated by using the formula:

v = ω.r

where

ω is angular velocity

r is radius of the path

<u>Angular</u> <u>velocity</u> is the rate of change of the angular position of an object with respect of time, i.e., is how fast an object changes its angular position with time:

$\omega = \frac{d \theta}{dt}$

$\omega = \frac{2\pi}{time}$

The Earth takes 24 hours to complete a rotation, then angular velocity:

$\omega = \frac{2\pi}{24}$

$\omega =$ 0.262 rad/h

At a location 40° north, radius is 3033.5 miles, so:

v = 0.262*3033.5

v = 793.77

<u>At latitude of 40°north, linear speed is </u><u>793.77mph</u>.

4 0

3 years ago

SOS if anyone could help that would be appreciated!!!❤️

Anastaziya [24]

Answer:

See below

Step-by-step explanation:

$2e)15-4x=10x+45$

Solve:

$15-4x=10x+45\\-4x-10x=45-15\\-14x=30\\ \boxed {x=-\dfrac{15}{7}}$ $3(x+1)=21$

Check:

$15-4 \left(-\dfrac{15}{7} \right)=10\left(-\dfrac{15}{7} \right)+45$

$15+\dfrac{60}{7} \right)=-\dfrac{150}{7} +45$

$\dfrac{210}{7} =30$

$30=30$

$3a) 3(x+1)=21$

Solve:

$3(x+1)=21\\3x+3=21\\3x=18\\x=6$

$3b) 2+3(x+1)=6x$

$2+3x+3=6x\\5=3x\\$

$x=\dfrac{5}{3}$

$3c) -2(2x-3)=-2$

Solve:

$-2(2x-3)=-2\\-4x+6=-2\\-4x=-8\\x=2$

3 0

3 years ago

Raj is simplifying (5^4)^3 using these steps:

Vika [28.1K]

$(a^{x})^{y}= a^{x*y}
\\ \\So,
(5^{4})^{3}= 5^{4*3} = 5^{12}$

7 0

3 years ago

Read 2 more answers