Help me for number 14 & 15

Vesnalui [34]

It's a flat horizontal line so the slope is 0

Slope = rise/run = 0/run = 0

The "run" can be any number you want as long as it's not 0.

4 0

3 years ago

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Can u help me plz its vocabulary

jeka57 [31]

Answer:

1.

Q- acute

R- right

S-acute

2.

- QRS is a right triangle because it has a small square at the right angle.

hope this helps!

Step-by-step explanation:

6 0

3 years ago

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The arithmetic sequence a; is defined by the formula:

Morgarella [4.7K]

Answer:

The sum of the first 650 terms of the given arithmetic sequence is 2,322,775

Step-by-step explanation:

The first term here is 4

while the nth term would be ai = a(i-1) + 11

Kindly note that i and 1 are subscript of a

Mathematically, the sum of n terms of an arithmetic sequence can be calculated using the formula

Sn = n/2[2a + (n-1)d)

Here, our n is 650, a is 4, d is the difference between two successive terms which is 11.

Plugging these values, we have

Sn = (650/2) (2(4) + (650-1)11)

Sn = 325(8 + 7,139)

Sn = 325(7,147)

Sn = 2,322,775

6 0

3 years ago

Raymond uses the Venmo

Arisa [49]

Answer:

2.29

Step-by-step explanation:

2.95 plus 1.26 is 4.21, and 6.50 minus 4.21 is 2.29.

5 0

3 years ago

A sanitation department is interested in estimating the mean amount of garbage per bin for all bins in the city. In a random sam

soldi70 [24.7K]

Answer:

a) $52.8-2.02\frac{3.9}{\sqrt{40}}=51.554$

b) $52.8+2.02\frac{3.9}{\sqrt{40}}=54.046$

c) $52.8-2.71\frac{3.9}{\sqrt{40}}=51.129$

d) $52.8+2.71\frac{3.9}{\sqrt{40}}=54.471$

e) Yes, depends of the confidence level.

At 95 % of confidence the value 54.1985 pounds is not included on the interval. At 5% of significance the statement is FALSE.

At 99 % of confidence the value 54.1985 pounds is included on the interval. So at 1% of significance the statement is TRUE.

Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X =52.8$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s=3.9 represent the sample standard deviation

n=40 represent the sample size

a) What is the lower limit of the 95% interval? Give your answer to three decimal places

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the mean and the sample deviation we can use the following formulas:

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=40-1=39$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,39)".And we see that $t_{\alpha/2}=2.02$

Now we have everything in order to replace into formula (1):

$52.8-2.02\frac{3.9}{\sqrt{40}}=51.554$

b) What is the upper limit of the 95% interval? Give your answer to three decimal places

$52.8+2.02\frac{3.9}{\sqrt{40}}=54.046$

So on this case the 95% confidence interval would be given by (51.554;54.046)

c) What is the lower limit of the 99% interval? Give your answer to three decimal places

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,39)".And we see that $t_{\alpha/2}=2.71$

Now we have everything in order to replace into formula (1):

$52.8-2.71\frac{3.9}{\sqrt{40}}=51.129$

d) What is the upper limit of the 99% interval? Give your answer to three decimal places

$52.8+2.71\frac{3.9}{\sqrt{40}}=54.471$

So on this case the 99% confidence interval would be given by (51.129;54.471)

e) Consider the claim that the mean amount of garbage per bin is 54.1985 pounds. Is the following statement true or false? The decision about the claim would depend on whether we use a 95% or 99% confidence interval: True/False

Yes, depends of the confidence level.

At 95 % of confidence the value 54.1985 pounds is not included on the interval. At 5% of significance the statement is FALSE.

At 99 % of confidence the value 54.1985 pounds is included on the interval. So at 1% of significance the statement is TRUE.

3 0

3 years ago