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3 years ago

240 projects in 15 days = projects per day

Mathematics

2 answers:

JulijaS [17]3 years ago

4 0

Answer:

16 projects per day

Step-by-step explanation:

Take the number of projects and divide by the number of days

240/15

16 projects per day

soldi70 [24.7K]3 years ago

3 0

Answer:

Step-by-step explanation:

240/15 is 16 projects per day

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I need help ASAP pls

Molodets [167]

Answer

Rise from the blue dot run to the red dot. Rise over run.

Step-by-step explanation:

1/8 would be your answer

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3 years ago

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Abigail mixes 2/3 pounds of walnuts with 3/5 pound of dried fruit. To create more of the same mixture, how many pounds of walnut

nignag [31]

1 1/9

To figure out how much walnuts to a pound of dried fruit you’d need to do proportions. So we cross multiply:

2/3 = X
—————- —————-
3/5 = 1

You multiply diagnols 2/3 x 1 = 2/3
3/5 times x is 3/5x.

2/3 = 3/5x. Isolate the variable. Divide 3/5 on both sides cancel out 3/5. 2/3 divide by 3/5 is 2/3 x 5/3 which is 10/9 which equals 1 1/9.

So for 1 lb of dried fruit you need 1 1/9 lb of walnuts to maintain the same mixture

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3 years ago

Which triangle is a translation of triangle x?

Vadim26 [7]

Answer:

Step-by-step explanation:

because x is 3 high and 2 across so is B

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3 years ago

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Need help with number 7

tatyana61 [14]

Since she runs 6 $\frac{1}{2}$ miles each day and there 7 days per week, you will have to multiply 6 $\frac{1}{2}$ times 7.

But since you have to use distributive properties, multiply like this (6 $\frac{1}{2}$ )(7)

6 $\frac{1}{2}$ is equal to $\frac{13}{2}$

Now turn 7 into a fraction, so its $\frac{7}{1}$

Now multipy (6 $\frac{1}{2}$ )( $\frac{7}{1}$ )

Multiply the numerators, so its 7 * 13 which gives you 91 and multiply the denominators so its 1 * 2 which gives you 2.

Put that back into a fraction as $\frac{91}{2}$

Divde it to change it to a mixed number and the answer is 45 $\frac{1}{2}$

8 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.

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3 years ago