Someone please help me

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]

Answer:

$159.99 * .25 "C"

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