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2 years ago

Convert 0.7135 to minutes and seconds

1 answer:

5 0

0.7135 minutes = 42.81 seconds
0.7135 seconds = 0.0118916667 seconds

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What is five billion two hundred fifty four million seventy one thousand nine hundred twenty six written in standard form

lakkis [162]

5,254,071,926

Is the standard form

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3 years ago

Use the diagram to find the measure of angle KHL.

bija089 [108]

Answer:

m∠KHL = 43°

Step-by-step explanation:

From the picture attached,

m∠KHL ≅ m∠GHL [Given in the picture]

Now substituting the values of the angles,

(3x + 1) = (5x - 27)

1 + 27 = 5x - 3x

28 = 2x

14 = x

m∠KHL = (3x + 1)° = (3 × 14) + 1

= 42 + 1

= 43°

Therefore, measure of ∠KHL = 43°

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3 years ago

the wildcats scores 24 more than twice the number of points that the stallions scored. Altogether, the two teams scored a total

Xelga [282]

Answer:

42?

Step-by-step explanation:

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3 years ago

Graph g(x) = –2x – 3

Luba_88 [7]

Answer:

The slope would be -2 with a y-intercept of (0, -3).

I hope this helped to answer your question.

4 0

2 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

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