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4 years ago

PLZ HELP YOU DON"T HAVE TO DO ALL BUT PLEASE SOME

Mathematics

2 answers:

Katarina [22]4 years ago

6 0

Question 1:
7/10= 0.7
71/100=0.71

Alecsey [184]4 years ago

3 0

3 it is the same
because do the division against the to decimals71.o and 9 divided by 1
2

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Which equation is in point-slope form and depicts the equation of this line?

crimeas [40]

Answer:

i believe the answer is A

Step-by-step explanation:

(y + 4) = -(1)/(3)(x + 1)

distribute -1/3 to (x+1)

(y+4)= -1/3x + -1/3

subtract 4 from each side

y= -1/3x - 13/3

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3 years ago

Which of these number lines represents the difference -3-2

Yuki888 [10]

Answer:

Number line D

Step-by-step explanation:

That line goes from 0 to -3 and subtracts 2, as the equation shows.

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3 years ago

Find the dot product of two given vectors, u = <1,6> and v = <-5,-2>

Crank

U.v = (1*-5) + (6*-2) = -5 - 12 = -17 <------- Answer

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3 years ago

The violent-crime rate in a certain state of the country in that year was 1,496. Would this be an outlier? O A. No, because it i

Kobotan [32]

Answer:

The violent-crime rate in a certain state of the country in that year was 1,496

Would this be an outlier?

C. Yes, because it is greater than the upper fence.

(d) Do you believe that the distribution of violent-crime rates is skewed or symmetric?

C. The distribution of violent-crime rates is skewed right.

Step-by-step explanation:

The violent-crime rate in a certain state of the country in that year was 1,496

The lower fence is 272.8 - 1.5 x 255.5 = -110.45 crimes per 100,000 population.

The upper fence is 528.3 + 1.5 x 255.5 = 911.55 crimes per 100,000 population.

Since violent-crime rate in this certain state of the country in that year is greater than the upper fence (1496>911.55), then it is an outlier.

(d) Do you believe that the distribution of violent-crime rates is skewed or symmetric?

The distribution of violent-crime rates is not symmetric, as there are extreme values in the tail, which tend to pull the mean in the direction of the tail to have data are either skewed left or skewed right, in this case it is skewed right, as there are large observations in the right tail that tend to increase the value of the mean, while having little effect on the median.

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3 years ago

Assuming that the population is normally distributed, construct a 9090% confidence interval for the population mean for each o

jasenka [17]

Given:

Set A: 1 4 4 4 5 5 5 8

Mean: 4.5

Standard dev: 1.9

Set B:

Mean: 4.5

Standard dev: 2.45

% = 90

Set A:

Standard Error, SE = s/ √n = 1.9/√8 = 0.67

Degrees of freedom = n - 1 = 8 -1 = 7

t- score = 1.89457861

Width of the confidence interval = t * SE = 1.89457861* 0.67 = 1.272685913

Lower Limit of the confidence interval = x-bar - width = 4.5 - 1.272685913 = 3.23

Upper Limit of the confidence interval = x-bar + width = 4.5 + 1.272685913 = 5.77

The 90% confidence interval is [3.23, 5.77]

Set B:

Standard Error, SE = s/ √n = 2.45/√8 = 0.87

Degrees of freedom = n - 1 = 8 -1 = 7

t- Score = 1.89457861

Width of the confidence interval = t * SE = 1.89457861* 0.87 = 1.641094994

Lower Limit of the confidence interval = x-bar - width = 4.5 - 1.641094994 = 2.86

Upper Limit of the confidence interval = x-bar + width = 4.5 + 1.641094994 = 6.14

The 90% confidence interval is [2.86, 6.14]

We can obviously see that sample B has more variation in the scores than sample A. The fact that the standard deviation is 2.45 for B and 1.9 for A). Therefore, they yield dissimilar confidence intervals even though they have the same mean and range.

6 0

4 years ago