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baherus [9]
3 years ago
8

The distance between 4.25 and 0 on a number line is the same distance as -4.24 units

Mathematics
1 answer:
Mariulka [41]3 years ago
3 0
Answer:

The distance between 0 and 4.25 is 4.25.

Solution: if you started from 1, it would be 4.24. And it will be positive because a distance can only be positive. You wouldn’t say you went walked -3 miles to your house, you would say you walked 3 miles to your house
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Find the slope of the line through the points -12, 10 and -9,-15
Pavlova-9 [17]

Answer:

-25/3

Step-by-step explanation:

7 0
3 years ago
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-y=3 would be y=-3 right?
tankabanditka [31]

Answer:

correct

Step-by-step explanation:

to make y positive, you would divide it by -1. This makes y positive and 3 negative.

3 0
3 years ago
Find the circumferences of both circles to the nearest tenth.
Liula [17]

Answer:

The Orange circle is: 78.5 cm

And The Blue circle is 314.2 cm

Step-by-step explanation:

By taking the formula r^2(pi), and filling it in with the radius for the orange circle you will get 78.5. For the blue circle the radius would be the orange's radius combined with the rim given. Using the same formula you will get 314.2.  

5 0
4 years ago
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Leroy Road six rides at the State Fair of Texas for $19.56 find constant of the proportionality
Flauer [41]
He rode 6 rides times some number that cost 19.56, therefore
6x =19.56
Solve for x to find what the cost is or in this case the value of proportion.
X=3.26
6 0
3 years ago
A group of students estimated the length of one minute without reference to a watch or​ clock, and the times​ (seconds) are list
Kitty [74]

Answer:

It does not appear ​that, as a​ group, the students are reasonably good at estimating one minute.

Step-by-step explanation:

We are given the following data in the question:

75, 88, 51, 73, 49, 31, 69, 74, 72, 59, 72, 81, 99, 101, 73

Formula:

\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}  

where x_i are data points, \bar{x} is the mean and n is the number of observations.  

Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}

Mean =\displaystyle\frac{1067}{15} = 71.13

Sum of squares of differences = 4739.733

S.D = \sqrt{\frac{4739.733}{14}} = 18.39

Population mean, μ = 60 minutes

Sample mean, \bar{x} = 71.13 minutes

Sample size, n = 15

Alpha, α = 0.10

Sample standard deviation, s = 18.39 minutes

First, we design the null and the alternate hypothesis

H_{0}: \mu = 60\text{ minutes}\\H_A: \mu \neq 60\text{ minutes}

We use Two-tailed t test to perform this hypothesis.

Formula:

t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }

Putting all the values, we have

t_{stat} = \displaystyle\frac{71.13 - 60}{\frac{18.39}{\sqrt{15}} } = 2.34

Calculating the p-value from the table, we have,

P-value = 0.034354

Since the p-value is lower than the significance level, we fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.

Thus, we conclude that it does not appear ​that, as a​ group, the students are reasonably good at estimating one minute.

7 0
3 years ago
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