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3 years ago

Find the percent of decrease from 340 to 160. Round to the nearest tenth of a

Mathematics

1 answer:

PIT_PIT [208]3 years ago

5 0

Answer: D is correct, 52.9%

Step-by-step explanation:

% decrease from 340 to 160, Where 340 is the old number and 160 is the new number

Percent change = (New number - Old number) / Old number x 100%

Percent change

= (160 - 340) / 340 x 100

= (-180) / 340x 100 = -52.94%

I hope this is clear, please mark as brainliest answer.

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What are the first three terms of the Arithmetic Sequence an=11-3(n-1)?

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Answer:

11,8,5

Step-by-step explanation:

an=11-3(n-1)

Let n=1

a1 = 11 - 3(1-1)

= 11 -0

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a2 = 11-3(2-1)

= 11 -3(1)

= 8

Let n=3

a3 = 11-3(3-1)

= 11 -3(2)

= 11 -6

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Explain how you can tell if the expressions

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Step-by-step explanation:

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3 years ago

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A salesperson is guaranteed $450 per week plus a 4% commission on all sales.

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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

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