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2 years ago

If one meter is approximately 3.28ft how many meters are in 20 ft

Mathematics

2 answers:

shusha [124]2 years ago

5 0

Answer:

I think the answer is 6.096 Meters.

Step-by-step explanation:

kotykmax [81]2 years ago

3 0

Answer:

10.76

Step-by-step explanation:

if you convert meters to feet

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You open a small business with an intial investment of $90,000. The weekly running costs for the business are $7800. the weekly

GREYUIT [131]

Answer:

90000=1000x

90 days

Step-by-step explanation:

90000=8800x-7800x

90000=1000x

4 0

2 years ago

Ava deposited $14 in a savings account that earns 2.7% simple interest. What would the graph look like?

katen-ka-za [31]

Deposited amount = $14.

Rate of interest = 2.7%

In decimals 2.7 could be written by dividing 2.7 by 100, we get

0.027.

Interest earned each year = 2.7% of $14 = 0.027*14 = $0.378.

We can make a table to balance by after each year.

First year balance: First year interest + Deposited amount = 0.378 +14 = $14.378.

Second year balance: Second year interest + Deposited amount = 2*0.378 +14 = 0.756+ 14 = $14.756.

Third year balance: Third year interest + Deposited amount = 3*0.378 +14 = 1.134+ 14 = $15.134.

We get four coordinates to graph A(0,14), B(1, 14.378) , C(2, 14.756) and D(3, 15.134).

Plotting those coordinates of the points A, B, C and D on the graph.

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

Help me please i would appreciate it alot (stop answering with things like 'im not sure' just put those in the comments)

OLEGan [10]

#3 would be positive

3 0

3 years ago

Read 2 more answers

How to find the area of the shaded region

Andreas93 [3]

Y=2x^(1/2)

A=2⌠x^(1/2) dx

A=2[ (2/3)x^(3/2) ]

A=(4/3)[ x^(3/2) ] x=[1,4]

A=(4/3)(8-1)

A=(4/3)7

A=28/3 u^2

A=9 1/3 u^2

7 0

3 years ago