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beks73 [17]
3 years ago
6

A tree grows the same amount each year.one year the tree is 48 inches tall.When it is measured 3 years later it is 84 inches.How

many inches does the tree grow each year?
Mathematics
1 answer:
slamgirl [31]3 years ago
8 0
84-48=36
36/3=12
The tree grows 12 inches every year
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Write a mathematical sentence that expresses the information given below.
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Answer:

73>16

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2/3 of a day (in hours) explain also​
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Can somebody help me with these 2 questions and show work ?? :)
777dan777 [17]

5. A. (4, -2)

6. C. (x, y) — (x, -y + 5)

Step-by-step explanation:

5. For the formula y = x, the x and y coordinates get swapped.

M = (-2, 4) — M’ = (4, -2)

6. If the coordinates get reflected across the x-axis, the y coordinates become negative.

(x, y) — (x, -y)

Now that the coordinates are reflected, you go 5 units up (+ 5) to get to the reflection of the coordinates if it was 5 units down before it reflected across the x-axis (- 5).

Ex. 1, 6 gets reflected across the x-axis and moved 5 units up. It’s reflection would be equivalent to (1, -1) because it moved 5 units down (1, 1) then reflected across the x-axis (1, -1).

(x, y - 5) reflected across the x-axis is equivalent to (x, -y + 5)

8 0
3 years ago
supervisor records the repair cost for 25 randomly selected dryers. A sample mean of $93.36 and standard deviation of $19.95 are
Simora [160]
<h2><u>Answer with explanation:</u></h2>

The confidence interval for population mean (when population standard deviation is unknown) is given by :-

\overline{x}-t^*\dfrac{s}{\sqrt{n}}< \mu

, where n= sample size

\overline{x} = Sample mean

s= sample size

t* = Critical value.

Given : n= 25

Degree of freedom : df=n-1=24

\overline{x}= \$93.36

s=\ $19.95

Significance level for 98% confidence interval : \alpha=1-0.98=0.02

Using t-distribution table ,

Two-tailed critical value for 98% confidence interval :

t^*=t_{\alpha/2,\ df}=t_{0.01,\ 24}=2.4922

⇒ The critical value that should be used in constructing the confidence interval = 2.4922

Then, the 95% confidence interval would be :-

93.36-(2.4922)\dfrac{19.95}{\sqrt{25}}< \mu

=93.36-9.943878< \mu

=93.36-9.943878< \mu

=83.416122< \mu

Hence, the 98% confidence interval for the mean repair cost for the dryers. = 83.4161

4 0
3 years ago
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