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3 years ago

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A matinee movie ticket costs $6.50 per child. Which equation can be used to calculate the matinee ticket price, M, for n childre

n? M = 6.50n n = 6.50M M equals 6.50 over n n equals 6.50 over M

2 answers:

emmainna [20.7K]3 years ago

8 0

Answer:

m = 6.50

Step-by-step explanation:

zloy xaker [14]3 years ago

5 0

Answer:

M = 6.50n

Step-by-step explanation:

since it is per child, it will be multiplied by the amount of children, n. Therefore giving us the cost of M, or the equivalent.

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Which equation is the inverse of y = x^2– 36?

Ierofanga [76]

Answer:

y inverse = √(x + 36)

Step-by-step explanation:

inverse of y = x² - 36

=> x² = y + 36

=> x = √(y + 36)

interchange x and y

=> y = √(x + 36)

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2 years ago

( −2 x^3) ^ 3 simplify

densk [106]

Answer:

$-8x^9$

Step-by-step explanation:

We can split up the exponent into multiple products: $(-2x^3)^3=(-2)^3*(x^3)^3$ .

We know that $(-2)^3=(-2)*(-2)*(-2)=-1*-1*-1*2*2*2=-1*8=-8$ .

We also know that $(x^3)^3=x^{3*3}=x^9$ .

So, our answer is $\boxed{-8x^9}$ and we're done!

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2 years ago

Read 2 more answers

The number of hours per day that Americans spend on social networking is approximately normally distributed. A random sample of

-Dominant- [34]

Answer:

B. (2.869, 3.411)

<em />

Step-by-step explanation:

<em>The question is incomplete:</em>

<em>The number of hours per day that Americans spend on social networking is approximately normally distributed. A random sample of 20 Americans who use social networks had M = 3.14 hours and s = 0.58 hours. Find a 95% confidence interval for the actual mean number of hours that Americans spend on social networking. </em>

<em> </em>

<em>A. (2.860, 3.420) </em>

<em> B. (2.869, 3.411) </em>

<em>C. (2.886, 3.394) </em>

<em>D. Not appropriate</em>

We have to calculate a 95% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=3.14.

The sample size is N=20.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

$s_M=\dfrac{s}{\sqrt{N}}=\dfrac{0.58}{\sqrt{20}}=\dfrac{0.58}{4.472}=0.13$

The degrees of freedom for this sample size are:

$df=n-1=20-1=19$

The t-value for a 95% confidence interval and 19 degrees of freedom is t=2.093.

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_M=2.093 \cdot 0.13=0.271$

Then, the lower and upper bounds of the confidence interval are:

$LL=M-t \cdot s_M = 3.14-0.271=2.869\\\\UL=M+t \cdot s_M = 3.14+0.271=3.411$

The 95% confidence interval for the mean is (2.869, 3.411).

3 0

3 years ago

How would you factor c(m - n) d(m - n)?

Kazeer [188]

Ab+ac=a(b+c) or
ba+ca=a(b+c)
c(m-n)+d(m-n)=(m-n)(c+d)

5 0

3 years ago

In studies for a medication, 3 percent of patients gained weight as a side effect. Suppose 643 patients are randomly selected.

timofeeve [1]

Part a)

It was given that 3% of patients gained weight as a side effect.

This means

$p = 0.03$

$q = 1 - 0.03 = 0.97$

The mean is

$\mu = np$

$\mu = 643 \times 0.03 = 19.29$

The standard deviation is

$\sigma = \sqrt{npq}$

$\sigma = \sqrt{643 \times 0.03 \times 0.97}$

$\sigma =4.33$

We want to find the probability that exactly 24 patients will gain weight as side effect.

P(X=24)

We apply the Continuity Correction Factor(CCF)

P(24-0.5<X<24+0.5)=P(23.5<X<24.5)

We convert to z-scores.

$P(23.5 \: < \: X \: < \: 24.5) = P( \frac{23.5 - 19.29}{4.33} \: < \: z \: < \: \frac{24.5 - 19.29}{4.33} ) \\ = P( 0.97\: < \: z \: < \: 1.20) \\ = 0.051$

Part b) We want to find the probability that 24 or fewer patients will gain weight as a side effect.

P(X≤24)

We apply the continuity correction factor to get;

P(X<24+0.5)=P(X<24.5)

We convert to z-scores to get:

$P(X \: < \: 24.5) = P(z \: < \: \frac{24.5 - 19.29}{4.33} ) \\ = P(z \: < \: 1.20) \\ = 0.8849$

Part c)

We want to find the probability that

11 or more patients will gain weight as a side effect.

P(X≥11)

Apply correction factor to get:

P(X>11-0.5)=P(X>10.5)

We convert to z-scores:

$P(X \: > \: 10.5) = P(z \: > \: \frac{10.5 - 19.29}{4.33} ) \\ = P(z \: > \: - 2.03)$

$= 0.9788$

Part d)

We want to find the probability that:

between 24 and 28, inclusive, will gain weight as a side effect.

P(24≤X≤28)=

P(23.5≤X≤28.5)

Convert to z-scores:

$P(23.5 \: < \: X \: < \: 28.5) = P( \frac{23.5 - 19.29}{4.33} \: < \: z \: < \: \frac{28.5 - 19.29}{4.33} ) \\ = P( 0.97\: < \: z \: < \: 2.13) \\ = 0.1494$

3 0

3 years ago