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3 years ago

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

1 answer:

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

Mama L [17]

Answer:

17 pieces of fruit in total

Step-by-step explanation:

2x7=14+3=17

7 0

3 years ago

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

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

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

Point D
$D(1,-2) \\ for x=1 \\ f(1)=-2$

therefore

the answer is
$f(0)=1$

4 0

3 years ago

Read 2 more answers

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

ASHA 777 [7]

Answer:

11.33 degrees

Step-by-step explanation:

5 0

2 years ago

Please help with 7 :))

Maru [420]

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.

Let $\mu$ = mean lead concentration for all such medicines.

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 One-sample t test statistics 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, test statistics = $\frac{12.25 -17}{\frac{6.96}{\sqrt{10}}}$ ~ $t_9$

= -2.16

The value of t test statistics is -2.16.

Now, the P-value of the test statistics is given by the following formula;

P-value = P( $t_9$ < -2.16) = 0.031

Now, at 0.10 significance level the t table gives critical value of -1.383 for left-tailed test. 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 we reject our null hypothesis.

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

7 0

3 years ago