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3 years ago

What is the image of the point

1 answer:

3 0

By estimation, 3.9 turns into 4 and 5.3 turns into 5. The answer after estimation is 20 while the real answer is 20.67. Well u see by estimation we get an answer of two places right. So, real answer should also have two places.Place the decimal point after it.y43

Step-by-step explanation: i dont know if I'm right or not

You might be interested in

Which number do you add to 3 to get 0?

grandymaker [24]

Answer:

-3

Step-by-step explanation:

3 + (-3) = 0

7 0

3 years ago

Read 2 more answers

A jet can carry up to 200,000 liters of fuel.it used 130,000 liters of fuel during a flight. What percentage of the fuel capacit

Komok [63]

Percentage=(fuel consumed/ full fuel capacity) *100%
=(130000/200000)*100%
=65%

3 0

2 years ago

Solve: x=4+(4x-4)1/2 x = 2

Andreyy89

Answer:

No real solution- Not any of the options anyway

Step-by-step explanation:

So I replaced the x's with the two possible equations and then tried to solve...

2=4+(4x2-4)1/2

2=4+(8-4)1/2

2=4+(4)1/2

2=8(1/2)

2=4 Which is not true

10=4+(4x10-4)1/2

10=4+(40-4)1/2

10=4+36(1/2)

10=40(1/2)

10=20 Which is also not true

Therefore, there is no real solution

6 0

3 years ago

Simplify. 9 – (–6) – 12

kondor19780726 [428]

9+6=15 and 15-12=3 so this =3 the -(-6) becomes positive

5 0

3 years ago

Read 2 more answers

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