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Lina20 [59]
3 years ago
14

A rate in which the two quantities are equal but different units is called a what

Mathematics
1 answer:
denis-greek [22]3 years ago
5 0

Answer: A rate with a second quantity of 1 unit. dimensional analysis. A process that uses rates to convert measurements from one unit to another. conversion factor. A rate in which the two quantities are equal but use different units is called this.


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Glenn ate 2 apples a day for a week. In addition to the apples, he ate 3 pears during the week. write the expression that shows
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Answer:

17 pieces of fruit in total

Step-by-step explanation:

2x7=14+3=17

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Use the graph below to fill in the blank with the correct number:
Alik [6]
Let
A(-3,0) \\ B(-2,2) \\ C(0,1) \\ D(1,-2)

we know that

Point A
A(-3,0) \\ for x=-3 \\ f(-3)=0

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B(-2,2) \\ for x=-2 \\ f(-2)=2

Point C
C(0,1) \\ for x=0 \\ f(0)=1

Point D
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Angle = (2x + 10)° and angle = (4y – 30)°. Find x and y.
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Answer:

11.33 degrees

Step-by-step explanation:

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2 years ago
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What Is Your. Answer
3 0
3 years ago
Listed below are the lead concentrations in mu ??g/g measured in different traditional medicines. Use a 0.10 0.10 significance l
laiz [17]

Answer:

We conclude that the mean lead concentration for all such medicines is less than 17 mu.

Step-by-step explanation:

We are given below are the lead concentrations in mu;

6.5 18.5 21.5 5.5 8.5 4.5 4.5 17.5 15.5 20

We have to test the claim that the mean lead concentration for all such medicines is less than 17 mu.

<u><em>Let </em></u>\mu<u><em> = mean lead concentration for all such medicines.</em></u>

So, Null Hypothesis, H_0 : \mu = 17 mu      {means that mean lead concentration for all such medicines is equal to 17 mu}

Alternate Hypothesis, H_A : \mu < 17 mu       {means that the mean lead concentration for all such medicines is less than 17 mu}

The test statistics that would be used here <u>One-sample t test</u> <u>statistics</u> as we don't know about population standard deviation;

                      T.S. =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n}}}  ~ t_n_-_1

where, \bar X = sample mean lead concentration = \frac{\sum X}{n} = 12.25 mu

             s = sample standard deviation = \sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} } = 6.96 mu

             n = sample size = 10

So, <u><em>test statistics</em></u>  =  \frac{12.25 -17}{\frac{6.96}{\sqrt{10}}}  ~ t_9

                              =  -2.16

The value of t test statistics is -2.16.

<u>Now, the P-value of the test statistics is given by the following formula;</u>

                P-value = P( t_9 < -2.16) = <u>0.031</u>

<u><em>Now, at 0.10 significance level the t table gives critical value of -1.383 for left-tailed test.</em></u><em> Since our test statistics is less than the critical value of t as -2.16 < -1.383, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which </em><em><u>we reject our null hypothesis</u></em><em>.</em>

Therefore, we conclude that the mean lead concentration for all such medicines is less than 17 mu.

7 0
3 years ago
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