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kotykmax [81]
3 years ago
6

5÷630 wrote out partial quotient

Mathematics
1 answer:
Cerrena [4.2K]3 years ago
5 0
126 is the answer becaus it is how stupid are you
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To increase the value of estimated Cohen's d, a researcher increases the sample size. Will this change the value of d?
tatyana61 [14]

Answer:

a. A) Yes; as sample size increases, effect size increases.

b. A) The critical value increases.

Step-by-step explanation:

a.

If all possible samples of size N are drawn from a finite population, Np, without replacement, and the standard deviation of the mean values of the sampling distribution of means is determined then:

σx =d  √N

d = σx . √N

d ∞ N

from this, we can say d is directly proportional to √N

where σx is the standard deviation of the sampling distribution of  means and d is the standard deviation of the population’. The standard deviation of a sampling distribution of mean values is called the standard error of the means,

therefore, we can conclude that:

A) Yes; as sample size increases, effect size increases.

b.

In this estimate, tc is called the confidence coefficient for small samples,  d is the standard deviation of the sample, x is the mean value of the sample and N is the number of members in the sample.

When determining the t-value, given by

t =( (x−µ) /s) * \sqrt{N-1}

it is necessary to know the sample parameters x and s and the population parameter µ. x and s can be calculated for the sample, but usually an estimate has to be made of the population mean µ, based on the sample mean value.

from the above equation it can be deduced that t value is determined with the sample size, and as the sample size increases

A) The critical value increases.

3 0
3 years ago
I need help plz and plz explain why plz
Mariana [72]

Answer:

156 minutes

Step-by-step explanation:

2 hrs=120 minutes

3/5 hr=3/5 *60=36 minutes

120+36=156 minutes

hope this helps

brainliest?

6 0
3 years ago
Read 2 more answers
Solve sin^2x+cos2x-1/4=0
sweet-ann [11.9K]
sin^{2}x + cos2x - \frac{1}{4} = 0
sin^{2}x + cos^{2}x - sin^{2}x = \frac{1}{4}
cos^{2}x = \frac{1}{4}
cos(x) = \pm \frac{1}{2}

x = \pm \frac{\pi}{3}
General solutions: x = \pi(n - \frac{1}{3}) and x = \pi(n + \frac{1}{3})
8 0
3 years ago
Jean has 11 trees in her yard and Anita has 8. Write the ratio of trees in Anita's yard to trees in Jean's yard
Brilliant_brown [7]

Answer:

The ratio is 11:8

Step-by-step explanation:

There's 11 trees in Jeans yard hence the first number of the ratio (11)

While Anita has 8, hence the second number in the ratio (8)

Therefore the ration of Jean's trees to Anita's is 11:8

8 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
Read 2 more answers
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