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cupoosta [38]
3 years ago
8

Someone please help me

Mathematics
1 answer:
cupoosta [38]3 years ago
4 0

Answer:

which one ? and which one Questions

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Find the y-intercept for each of the new equations:
Zepler [3.9K]

Answer:whgdwyhdiuwojm

Step-by-step explanation:

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7 0
2 years ago
Explain how to compare 513251 and 513521 to find the greater number
Annette [7]
What you do is look at the numbers in order from the greatest number, to the least. For example, in this problem, we would start with the 5 in 513251, and the 5 in 513521. Since they are both fives, we don't know yet which one is bigger. So we move onto the next number, which would be the 1 in 513251, and the 1 in 513521. They are both the same, so we go onto the 3 in 513251, and the 3 in 513521. Those are also both the same, to we go onto the 2 in 513251, and the 5 in 513521. As you can see, the 5 from 513521 is greater then the 2 in 513251, so that shows that 513521 is greater.
6 0
3 years ago
Please help this one doesnt nee and explantion
ICE Princess25 [194]

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Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
T is the midpoint of KL. K has coordinates (2,−6), and T has coordinates (−4,2). Identify the coordinates of L.
kotykmax [81]

Answer:

L = (-10,10)

Step-by-step explanation:

Given

K = (2,-6)

T = (-4,2)

Required

Determine the coordinates of L

Since T is the midpoint of K and L, we make use of:

T_x = \frac{K_x + L_x}{2}

and

T_y = \frac{K_y + L_y}{2}

Solving for L_x

T_x = \frac{K_x + L_x}{2}

-4 = \frac{2 + L_x}{2}

Multiply through by 2

-8 = 2 + L_x

L_x = -8 - 2

L_x = -10

Solving for L_y

T_y = \frac{K_y + L_y}{2}

2 = \frac{-6+L_y}{2}

Multiply through by 2

4 = -6 + L_y

L_y = 4 + 6

L_y = 10

Hence: The coordinates of L is:

L = (-10,10)

6 0
3 years ago
Suppose twenty-two communities have an average of = 123.6 reported cases of larceny per year. assume that σ is known to be 36.8
Delvig [45]
We are given the following data:

Average = m = 123.6
Population standard deviation = σ= psd = 36.8
Sample Size = n = 22

We are to find the confidence intervals for 90%, 95% and 98% confidence level.

Since the population standard deviation is known, and sample size is not too small, we can use standard normal distribution to find the confidence intervals.

Part 1) 90% Confidence Interval
z value for 90% confidence interval = 1.645

Lower end of confidence interval = m-z *\frac{psd}{ \sqrt{n} }
Using the values, we get:
Lower end of confidence interval=123.6-1.645* \frac{36.8}{ \sqrt{22}}=110.69

Upper end of confidence interval = m+z *\frac{psd}{ \sqrt{n} }
Using the values, we get:
Upper end of confidence interval=123.6+1.645* \frac{36.8}{ \sqrt{22}}=136.51

Thus the 90% confidence interval will be (110.69, 136.51)

Part 2) 95% Confidence Interval
z value for 95% confidence interval = 1.96

Lower end of confidence interval = m-z *\frac{psd}{ \sqrt{n} }
Using the values, we get:
Lower end of confidence interval=123.6-1.96* \frac{36.8}{ \sqrt{22}}=108.22

Upper end of confidence interval = m+z *\frac{psd}{ \sqrt{n} }
Using the values, we get:
Upper end of confidence interval=123.6+1.96* \frac{36.8}{ \sqrt{22}}=138.98

Thus the 95% confidence interval will be (108.22, 138.98)

Part 3) 98% Confidence Interval
z value for 98% confidence interval = 2.327

Lower end of confidence interval = m-z *\frac{psd}{ \sqrt{n} }
Using the values, we get:
Lower end of confidence interval=123.6-2.327* \frac{36.8}{ \sqrt{22}}=105.34
Upper end of confidence interval = m+z *\frac{psd}{ \sqrt{n} }
Using the values, we get:
Upper end of confidence interval=123.6+2.327* \frac{36.8}{ \sqrt{22}}=141.86

Thus the 98% confidence interval will be (105.34, 141.86)


Part 4) Comparison of Confidence Intervals
The 90% confidence interval is: (110.69, 136.51)
The 95% confidence interval is: (108.22, 138.98)
The 98% confidence interval is: (105.34, 141.86)

As the level of confidence is increasing, the width of confidence interval is also increasing. So we can conclude that increasing the confidence level increases the width of confidence intervals.
3 0
3 years ago
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