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Reika [66]
3 years ago
7

Eckhart school is selling candy to raise money for its sports programs . Last year the school raised $480 during a 12 day fundra

iser. The expects this years daily sales to be in proportion to last years daily sales . Based on the schools expectations ,which of the following statement is true ?
Mathematics
2 answers:
Galina-37 [17]3 years ago
8 0

Answer:

120

Step-by-step explanation:

:)

Ber [7]3 years ago
3 0

Answer:

answer my question

Step-by-step explanation:

help pls!!!!!

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Three ratios that are Equivalent to 11:1
marin [14]

Answer:

1-22:2

2-33:3

3-44:4

Step-by-step explanation:

11+11=22

22+11=33

33+11=44

1+1=2

2+1=3

3+1=4

so, 22:2

33:3

44:4

6 0
3 years ago
How to work out 215 in a fraction
ohaa [14]
\frac{215}{215}
5 0
4 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
Rewrite 4/7 and 3/5 so they have common denominator
Vladimir [108]
20/35 and 21/35. The common denominator is 35.
6 0
3 years ago
What is 965,000,000,000,000 in scientific notation? 
stellarik [79]
The answer is 9.65x10^14 because in order for the decimal point to be between 9 and 6 you need to move 14 decimal places back
4 0
4 years ago
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