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Tresset [83]
2 years ago
10

Mr. Sonny's science class is calculating the average number of blinks per minute. Jan blinks 126 times in 7 minutes. What is her

blinking rate, in blinks per minute?
Mathematics
2 answers:
vlabodo [156]2 years ago
3 0
18 blinks per min since 126 divided by 7 equals 18
const2013 [10]2 years ago
3 0
Answer:
b. Every 24 minutes
Step-by-step explanation:
A neon sign blinks every 6 minutes, and another sign blinks every 8 minutes.
The first step is to:
Find and list multiples of each time each neon sign blinks.
This means, we find the Multiples of 6 and 8
Multiples of 6:
6, 12, 18, 24, 30, 36
Multiples of 8:
8, 16, 24, 32, 40
The first and common multiple of 6 and 8.
Therefore, the Least common multiple [LCM(6, 8) ] = 24
Hence, they often blink at the same time every 24 minutes.
Therefore, option b.Every 24 minutes is correct.
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Find the mean and median of these data: 2, 5, 13, 15, 19, 21.
Maru [420]

The Arithmetic Mean and Median of the given set of data ( 2, 5, 13, 15, 19, 21 ) are 12.5 and 14 respectively.

<h3>What is Arithmetic mean?</h3>

Arithmetic mean is simply the average of a given set numbers. It is determined by dividing the sum of a given set number by their number of appearance.

Mean = Sum total of the number ÷ n

Where n is number of numbers

Median is the middle number in the data set.

Given the sets;

  • 2, 5, 13, 15, 19, 21
  • n = 6

Mean = Sum total of the number ÷ n

Mean = (2 + 5 + 13 + 15 + 19 + 21) ÷ 6

Mean = 75 ÷ 6

Mean = 12.5

Median is the middle number in the data set.

Median = ( 13 + 15 ) ÷ 2

Median = 14

Therefore, the Arithmetic Mean and Median of the given set of data ( 2, 5, 13, 15, 19, 21 ) are 12.5 and 14 respectively.

Learn more about arithmetic mean here: brainly.com/question/13000783

#SPJ1

3 0
1 year ago
A sample of gold has a mass of 579 g. The volume of the sample is 30 cm3. What is the density of the gold sample?
Ugo [173]

The density of the gold sample is 19.3g/cm³

Let the mass of the gold sample be represented by m

m = 579 g

Let the volume of the gold sample be represented by V

V = 30 cm³

Let the density of the gold sample be represented by ρ

The formula for the density is:

Density = \frac{Mass}{Volume}

\rho = \frac{m}{V} \\\rho = \frac{579}{30} \\\rho = 19.3 g/cm^3

The density of the gold sample is 19.3g/cm³

Learn more here: brainly.com/question/17780219

3 0
2 years ago
Layana’s house is located at (2 and two-thirds, 7 and one-third) on a map. The store where she works is located at (–1 and one-t
NeTakaya

We have been given that Layana’s house is located at (2\frac{2}{3}, 7\frac{1}{3}) on a map. The store where she works is located at (-1\frac{1}{3}, 7\frac{1}{3}).

We are asked to find the distance from Layana’s home to the store

We will use distance formula to solve our given problem.

Let us convert our given coordinates in improper fractions.

2\frac{2}{3}\Rightarrow \frac{8}{3}

7\frac{1}{3}\Rightarrow \frac{22}{3}

-1\frac{1}{3}\Rightarrow -\frac{4}{3}

Now we will use distance formula to solve our given problem.

D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Upon substituting coordinates of our given point in above formula, we will get:

D=\sqrt{(\frac{22}{3}-\frac{22}{3})^2+(\frac{8}{3}-(-\frac{4}{3}))^2}

D=\sqrt{(0)^2+(\frac{8}{3}+\frac{4}{3})^2}

D=\sqrt{0+(\frac{8+4}{3})^2}

D=\sqrt{(\frac{12}{3})^2}

D=\sqrt{(4)^2}

D=4

Therefore, the distance from Layana's home to the store is 4 units and option A is the correct choice.

3 0
3 years ago
Read 2 more answers
Please help me solve this problem
Alex
The answer for this is X+2g= 0
8 0
2 years ago
Last year, there were 120 students in the<br> marching band. This year, there are 138.
Marrrta [24]

Answer:

dude nice but how is this a question

Step-by-step explanation:

CHARLI VAPES

NO TALENT

IM JUST 16

7 0
3 years ago
Read 2 more answers
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