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3 years ago

Solve the system by graphing. Write the solution as an ordered pair.

Mathematics

1 answer:

storchak [24]3 years ago

8 0

Answer:

the answer to the question is - 2 over 7 and - 12 over 7

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What expression can be used to add 7/10 + 13/20

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0.7+0.65 I’m not sure this is what u were looking for but I hope it helps have a good one

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3 years ago

Teacher has 1p she buys 10 boxes worth 10p how much does she have to pay

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3 years ago

Solve for r. 0.5(r+2.75) = 3

nirvana33 [79]

Answer:

r=3.25

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

The sum of two numbers is less than 2. If we subtract the second number from the first, the difference is greater than 1. What a

Anna007 [38]

Answer:

the value of these two numbers are 1/2 and 3/2

Step-by-step explanation:

Given that:

x+y < 2 ---- (1)

y - x > 1 ---- (2)

From equation (2), let y > 1 + x then substitute it into equation (1)

x + 1 + x < 2

2x + 1 < 2

2x < 2 - 1

2x < 1

x < 1/2

From equation (2), replace the value of x to be 1/2

y - 1/2 > 1

y > 1 + 1/2

y > 3/2

5 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