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3 years ago

A gray crayon is made with 5 ml of black wax for every 6 mL of white wax

Mathematics

2 answers:

Nikitich [7]3 years ago

8 0

B because 18 isn’t much bigger than 15. I guessed so I am more than likely wrong

german3 years ago

4 0

Answer:

B and C

Step-by-step explanation:

Reduce each ratio down to it's simplest form.

Black:White

A. 10:24 = 5:12

B. 15:18 = 5:6

C. 35:42 = 5:6

D. 20:26 = 10:13

E. 45:48 = 15:16

B and C have the same ratio, so they will create the and shade.

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Devon has 14 steel balls of equal weight. If he puts 8 of them in one pan of a balance, and the rest along with a weight of 20 g

astra-53 [7]

Let n be a number of balls: n = 14

In one pan you have 8 balls: 8n
In the other pan you have the rest of balls (14 - 8 = 6) along with a <span>weight of 20 grams = 6n + 20
So the pans are balanced:
8n = 6n + 20

Now, just solve the equation:
8n - 6n = 20
2n = 20
n = 20/2
n = 10
So, you know that a ball weights 10 grams.</span>

4 0

3 years ago

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What 3d shape does this

natima [27]

Triangular prism will be your answer

4 0

3 years ago

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Find f(-2) if f(x)=4x^2+3x-5

valentina_108 [34]

Answer:

f(-2) = 5

Step-by-step explanation:

To find f(-2), plug in -2 wherever there is an x.

f(x) = 4x² + 3x - 5

f(-2) = 4(-2)² + 3(-2) - 5

f(-2) = 4(4) + 3(-2) - 5

f(-2) = 16 - 6 - 5

f(-2) = 10 - 5

f(-2) = 5

3 0

3 years ago

Aisha organize her books for 1/6 hour and her coin collection for 3/7 hour what is the best estimate for the total time Aisha sp

jeyben [28]

This is the ans plus u can use maths from Microsoft app for these type of questions

4 0

3 years ago

On the first of each month, Shelly runs a 5k race. She keeps track of her times to track her progress. Her time in minutes is re

Vikki [24]

Answer:

C . 0.76

Step-by-step explanation:

We are given that on the first of each month , Shelly runs a 5 k race. She keeps track of her times to track her progress. Her time in minutes is recorded in the table:

Jan 40.55 Feb 41.51

March 42.01 Apr 38.76

May 36.32 June 34.28

July 35.38 Aug 37.48

Sept 40.87 Oct 48.32

Nov 41.59 Dec 42.71

We have to find the difference between the mean of the data , including the outlier and excluding the outlier

Outlier: That observation which is different from other observations.

The outlier in the given observations is 48.32 because is different from other observations.

Mean of the data including the outlier

Mean = $\frac{Sum \;of\;observations}{Total\;number\;of\;observations}$

Mean= $\frac{40.55+41.51+42.01+38.76+36.32+34.28+35.38+37.48+40.87+48.32+41.59+42.71}{12}$

Mean= $\frac{479.78}{12}$

Mean=39.982

Mean of the data excluding the outlier

Mean= $\frac{40.55+41.51+42.01+38.76+36.32+34.28+35.38+37.48+40.87+41.59+42.71}{11}Mean=[tex]\frac{431.46}{11}$

Mean=39.224

Difference between mean of the data including the outlier and excluding the outlier=39.982-39.224=0.758

Difference between mean of the data including the outlier and excluding the outlier=0.76

Answer Answer:

Step-by-step explanation:

We are given that on the first of each month , Shelly runs a 5 k race. She keeps track of her times to track her progress. Her time in minutes is recorded in the table:

Jan 40.55 Feb 41.51

March 42.01 Apr 38.76

May 36.32 June 34.28

July 35.38 Aug 37.48

Sept 40.87 Oct 48.32

Nov 41.59 Dec 42.71

We have to find the difference between the mean of the data , including the outlier and excluding the outlier

Outlier: That observation which is different from other observations.

The outlier in the given observations is 48.32 because is different from other observations.

Mean of the data including the outlier

Mean = $\frac{Sum \;of\;observations}{Total\;number\;of\;observations}$

Mean= $\frac{40.55+41.51+42.01+38.76+36.32+34.28+35.38+37.48+40.87+48.32+41.59+42.71}{12}$

Mean= $\frac{479.78}{12}$

Mean=39.982

Mean of the data excluding the outlier

Mean= $\frac{40.55+41.51+42.01+38.76+36.32+34.28+35.38+37.48+40.87+41.59+42.71}{11}Mean=[tex]\frac{431.46}{11}$

Mean=39.224

Difference between mean of the data including the outlier and excluding the outlier=39.982-39.224=0.758

Difference between mean of the data including the outlier and excluding the outlier=0.76

Answer :C . 0.76

5 0

3 years ago

Read 2 more answers