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3 years ago

7

If you were to use the substitution method to solve the following system, choose the new equation after the expression equivalen

t to x from the second equation is substituted into the first equation.
2x − 3y = −29
x + 4y = 13

A) 2(4y + 13) − 3y = −29
B) 2(−4y + 13) − 3y = −29
C) 2x − 3(4y + 13) = −29
D) 2x − 3(−4y + 13) = −29

1 answer:

Savatey [412]3 years ago

4 0

I’m pretty sure it’s A

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Answer:

4 batches: $2\frac{2}{3}$ cups

7 batches: $4\frac{2}{3}$ cups

Step-by-step explanation:

If he uses 2/3 cup of flour for <u>each</u> batch, then that is 2/3 four times, or 2/3 · 4.

$\frac{2}{3}$ × $\frac{4}{1}$

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$\frac{2}{3}$ × $\frac{3}{1}$

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Thus, $2\frac{2}{3}$ and $4\frac{2}{3}$ are the answers.

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1 year ago

Can you tell how many students got allowances of $15? Explain.

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2 years ago

Simplify the number into simplest radical form. Use the factor tree to help determine the factors. StartRoot 96 Endroot StartRoo

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Answer:

$[tex]4608\sqrt{3}$ [/tex]

Step-by-step explanation:

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2. $\sqrt{2^{5} *3} \sqrt{6} *2\sqrt{6}*4\sqrt{6}*4\sqrt{3}$ <em>factoring 96</em>

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3. $\sqrt{2^{5} } \sqrt{3}\sqrt{6} *2\sqrt{6}*4\sqrt{6}*4\sqrt{3}$

<em>using exponent rule - $(a^{b}) ^{c} = a^{bc}$ </em>

<em> $\sqrt{2^{5} } = 2^{5/2}$ </em>

4. $2^{5/2}\sqrt{3}\sqrt{2*3} *2\sqrt{6}*4\sqrt{6}*4\sqrt{3}$

<em>doing some simple simplification and $4=2^{2}$ and 6=2*3</em>

5. $2^{5/2} \sqrt{3} \sqrt{2} \sqrt{3} *2\sqrt{2} \sqrt{3} *2^{2}\sqrt{2} \sqrt{3}*4\sqrt{3}$

<em>collecting the roots on one side and applying exponent rule</em>

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<em>combining all powers of 2</em>

8. $2^{1/2+1/2+1/2+5/2+1+2+2}*3^{1/2+1/2+1/2+1/2+1/2}$

<em>Simplifying</em>

9. $2^{9} *3^{2}\sqrt{3}$

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11. $4608\sqrt{3}$

3 0

3 years ago

Read 2 more answers

The mean annual cost of an automotive insurance policy is normally distributed with a mean of $1140 and standard deviation of $3

DerKrebs [107]

Using the normal distribution, it is found that the probabilities are given as follows:

a) 0.8871 = 88.71%.

b) 0.0778 = 7.78%.

c) 0.8485 = 84.85%.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

$Z = \frac{X - \mu}{\sigma}$

The z-score measures how many standard deviations the measure is above or below the mean.

Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

The parameters in this problem are given as follows:

$\mu = 1140, \sigma = 310, n = 16, s = \frac{310}{\sqrt{16}} = 77.5$

Item a:

The probability is the <u>p-value of Z when X = 1250 subtracted by the p-value of Z when X = 1000</u>, hence:

X = 1250:

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{1250 - 1140}{77.5}$

Z = 1.42

Z = 1.42 has a p-value of 0.9222.

X = 1000:

$Z = \frac{X - \mu}{s}$

$Z = \frac{1000 - 1140}{77.5}$

Z = -1.81

Z = -1.81 has a p-value of 0.0351.

0.9222 - 0.0351 = 0.8871 = 88.71% probability.

Item b:

The probability is <u>one subtracted by the p-value of Z when X = 1250</u>, hence:

1 - 0.9222 = 0.0778 = 7.78%.

Item c:

The probability is the <u>p-value of Z when X = 1220</u>, hence:

$Z = \frac{X - \mu}{s}$

$Z = \frac{1220 - 1140}{77.5}$

Z = 1.03

Z = 1.03 has a p-value of 0.8485.

0.8485 = 84.85% probability.

More can be learned about the normal distribution at brainly.com/question/4079902

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3 0

2 years ago

Lily weighs 57.7 kg while her sister weighs 59.02 kg, how much more does Lily's sister weigh then her? Please provide the calcul

PilotLPTM [1.2K]

The answer is 1.32

For this calculation, assuming that no conversions are required, the calcitonin goes as follows:

All that is needed is to subtract 57.7 from 59.02.

59.02 - 57.7 = ?

59.02 - 57.7 = 1.32

Hope this helps! Have a great day.

5 0

2 years ago