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2 years ago

8

on Saturday 12 friends go ice skating altogether they pay $83.40 for admission they share the cost equally how much does each pe

rson pay

2 answers:

jenyasd209 [6]2 years ago

7 0

$6.95
per person
divide cost by number of people

Vladimir79 [104]2 years ago

3 0

$6.95 each... just divide 83.40 and 12 and you got the answer :)

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Suppose a baker claims that the average bread height is more than 15cm. Several of this customers do not believe him. To persuad

laila [671]

Answer:

$17-2.262\frac{1.9}{\sqrt{10}}=15.641$

$17+2.262\frac{1.9}{\sqrt{10}}=18.359$

So on this case the 95% confidence interval would be given by (15.641;18.359)

And since the lower limit for the confidence interval is higher than 15 we can conclude that at 5% of significance the true mean is higher than 15 cm

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=17$ represent the sample mean

$\mu$ population mean (variable of interest)

s=1.9 represent the sample standard deviation

n=10 represent the sample size

Confidence interval

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=10-1=9$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,9)".And we see that $t_{\alpha/2}=2.262$

Now we have everything in order to replace into formula (1):

$17-2.262\frac{1.9}{\sqrt{10}}=15.641$

$17+2.262\frac{1.9}{\sqrt{10}}=18.359$

So on this case the 95% confidence interval would be given by (15.641;18.359)

And since the lower limit for the confidence interval is higher than 15 we can conclude that at 5% of significance the true mean is higher than 15 cm

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3 years ago

Pls help me that would be great

omeli [17]

No. No. It is not linear because the slope has to be able to be plugged in to any x, y coordinates and be placed on the line.

4 0

3 years ago

I need help ASAP can someone please help me right away

zysi [14]

Answer:

Smallest number would be 1^o

Step-by-step explanation:

Now with this one I can see the question now and the smallest* number of degrees to tilt it would be 1 degrees.

Have a good day. :)

Hope this helps!

<u>(brainliest would be appreciated)</u>

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3 years ago

Lorico [155]

C. Jasmine bought some tiles to put around her patio. . .

They asked for area which is always L x W, which was the give away.

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3 years ago

A professor pays 25 cents for each blackboard error made in lecture to the student who pointsout the error. In a career ofnyears

marta [7]

Answer:

(a) The probability that <em>Y</em>₂₀ exceeds 1000 is 3.91 × 10⁻⁶.

(b) <em>n</em> = 28.09

Step-by-step explanation:

The random variable <em>Y</em>ₙ is defined as the total numbers of dollars paid in <em>n</em> years.

It is provided that <em>Y</em>ₙ can be approximated by a Gaussian distribution, also known as Normal distribution.

The mean and standard deviation of <em>Y</em>ₙ are:

$\mu_{Y_{n}}=40n\\\sigma_{Y_{n}}=\sqrt{100n}$

(a)

For <em>n</em> = 20 the mean and standard deviation of <em>Y</em>₂₀ are:

$\mu_{Y_{n}}=40n=40\times20=800\\\sigma_{Y_{n}}=\sqrt{100n}=\sqrt{100\times20}=44.72\\$

Compute the probability that <em>Y</em>₂₀ exceeds 1000 as follows:

$P(Y_{n}>1000)=P(\frac{Y_{n}-\mu_{Y_{n}}}{\sigma_{Y_{n}}}>\frac{1000-800}{44.72})\\=P(Z> 4.47)\\=1-P(Z$

**Use a <em>z </em>table for probability.

Thus, the probability that <em>Y</em>₂₀ exceeds 1000 is 3.91 × 10⁻⁶.

(b)

It is provided that P (<em>Y</em>ₙ > 1000) > 0.99.

$P(Y_{n}>1000)=0.99\\1-P(Y_{n}$

The value of <em>z</em> for which P (Z < z) = 0.01 is 2.33.

Compute the value of <em>n</em> as follows:

$z=\frac{Y_{n}-\mu_{Y_{n}}}{\sigma_{Y_{n}}}\\2.33=\frac{1000-40n}{\sqrt{100n}}\\2.33=\frac{100}{\sqrt{n}}-4\sqrt{n} \\2.33=\frac{100-4n}{\sqrt{n}} \\5.4289=\frac{(100-4n)^{2}}{n}\\5.4289=\frac{10000+16n^{2}-800n}{n}\\5.4289n=10000+16n^{2}-800n\\16n^{2}-805.4289n+10000=0$

The last equation is a quadratic equation.

The roots of a quadratic equation are:

$n=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

a = 16

b = -805.4289

c = 10000

On solving the last equation the value of <em>n</em> = 28.09.

8 0

3 years ago