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3 years ago

shen baked b cookies. his family ate 11 of them. using b, write an expression for the number of cookies that remained.

Mathematics

1 answer:

Jobisdone [24]3 years ago

7 0

Step-by-step explanation:

So, lets write out the equation you have right now.

B - 11 = ?

You can have B be whatever you want. It must be higher than 11 (unless your teacher or whoever your guardian is allows negative numbers). In this case, I will use 15.

B = 15

15 - 11 = ?

Solve the equation and you get 4. Simple!

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tankabanditka [31]

This is wrong.

It's supposed to be -4/15.

But try -4/5.

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Answer:

yes

Step-by-step explanation:

A rational number is in the form p/q where p and q are both integers and the denominator q is not equal to zero. This means that the set of all real numbers are rational numbers.

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What is an equation of the line, in point-slope form, that passes through (8, -8); slope: 3

gavmur [86]

Answer:

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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

Which equation shows y+15=3x in standard form?

Morgarella [4.7K]

Rearranging y + 15 = 3x we have
3x - y = 15

so its none of them

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