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Softa [21]
2 years ago
7

Please I need your help on all the answers!

Mathematics
1 answer:
lapo4ka [179]2 years ago
6 0

Answer: That’s all youre going to search

Step-by-step explanation:

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Ben claims that 12 is a factor of 24.how can you check whether he is correct?
Anna71 [15]

Divide the  24  by Ben's  12 . 

If his  12  is a factor, then the division will come out even,
and the answer will be a whole number.

A).  6 x 4 = 24

B).  5 x 9 = 45

C).  3 x 8 = 24

D).  6 x 9 = 54

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2 years ago
Point L is on line segment
KengaRu [80]

Answer:

k is the option that makes sense

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2 years ago
In the diagram, the radius of the outer circle is 2x cm and
hodyreva [135]

Area shaded = Area big circle- Area of small circle;

200 pi= pi•(2x)^2 -pi•6^2;

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200pi=pi•4(x^2 -9) divide both sides by 4pi;

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8 0
3 years ago
A quality analyst of a tennis racquet manufacturing plant investigates if the length of a junior's tennis racquet conforms to th
Burka [1]

Answer:

Confidence interval : 21.506 to 24.493

Step-by-step explanation:

A quality analyst selects twenty racquets and obtains the following lengths:

21, 25, 23, 22, 24, 21, 25, 21, 23, 26, 21, 24, 22, 24, 23, 21, 21, 26, 23, 24

So, sample size = n =20

Now we are supposed to find Construct a 99.9% confidence interval for the mean length of all the junior's tennis racquets manufactured at this plant.

Since n < 30

So we will use t-distribution

Confidence level = 99.9%

Significance level = α = 0.001

Now calculate the sample mean

X=21, 25, 23, 22, 24, 21, 25, 21, 23, 26, 21, 24, 22, 24, 23, 21, 21, 26, 23, 24

Sample mean = \bar{x}=\frac{\sum x}{n}

Sample mean = \bar{x}=\frac{21+25+23+22+24+21+25+21+23+ 26+ 21+24+22+ 24+23+21+ 21+ 26+23+ 24}{20}

Sample mean = \bar{x}=23

Sample standard deviation = \sqrt{\frac{\sum(x-\bar{x})^2}{n-1}}

Sample standard deviation = \sqrt{\frac{(21-23)^2+(25-23)^2+(23-23)^2+(22-23)^2+(24-23)^2+(21-23)^2+(25-23)^2+(21-23)^2+(23-23)^2+(26-23)^2+(21-23)^2+(24-23)^2+(22-23)^2+(24-23)^2+(23-23)^2+(21-23)^2+(21-23)^2+(26-23)^2+(23-23)^2+(24-23)^2}{20-1}}

Sample standard deviation= s = 1.72

Degree of freedom = n-1 = 20-1 -19

Critical value of t using the t-distribution table t_{\frac{\alpha}{2} = 3.883

Formula of confidence interval : \bar{x} \pm t_{\frac{\alpha}{2}} \times \frac{s}{\sqrt{n}}

Substitute the values in the formula

Confidence interval : 23 \pm 1.73 \times \frac{1.72}{\sqrt{20}}

Confidence interval : 23 -3.883 \times \frac{1.72}{\sqrt{20}} to 23 + 3.883 \times \frac{1.72}{\sqrt{20}}

Confidence interval : 21.506 to 24.493

Hence Confidence interval : 21.506 to 24.493

3 0
3 years ago
Read 2 more answers
THIS IS THE PICTURES TO MY OTHER QUESTION PLEASE ANSWER!!!! TELL ME IF ITS CORRECT AND GIVE ME AN EXPLANATION
nikklg [1K]

Answer:

skp

Step-by-step explanation:

7 0
2 years ago
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