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2 years ago

There are 3/20 tickets left to sell and capacity is 3000 how many tickets are available to sell

Mathematics

1 answer:

Musya8 [376]2 years ago

5 0

Answer:

Step-by-step explanation

20x = 900

----- -------

20 20

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PLS HELP ASAP THANKS ILL GIVE BRAINLKEST PLS PLS HELP ASAP ILL GIVE BRAINLKEST PLS

faltersainse [42]

Answer:

first one is 42/7 = 6

11*6 = 66

second one

1/2 inch = 3/2 feet

17/2 inches = 17*3 = 51/2 = 25.5

8 0

2 years ago

Read 2 more answers

An economist uses the price of a gallon of milk as a measure of inflation. She finds that the average price is $3.82 per gallon

salantis [7]

Answer:

(a) The standard error of the mean in this experiment is $0.052.

(b) The probability that the sample mean is between $3.78 and $3.86 is 0.5587.

(c) The probability that the difference between the sample mean and the population mean is less than $0.01 is 0.5754.

(d) The likelihood that the sample mean is greater than $3.92 is 0.9726.

Step-by-step explanation:

According to the Central Limit Theorem if we have an unknown population with mean μ and standard deviation σ and appropriately huge random samples (n > 30) are selected from the population with replacement, then the distribution of the sample means will be approximately normally distributed.

Then, the mean of the distribution of sample mean is given by,

$\mu_{\bar x}=\mu$

And the standard deviation of the distribution of sample mean is given by,

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}$

The information provided is:

$n=40\\\mu=\$3.82\\\sigma=\$0.33$

As n = 40 > 30, the distribution of sample mean is $\bar X\sim N(3.82,\ 0.052^{2})$ .

(a)

The standard error is the standard deviation of the sampling distribution of sample mean.

Compute the standard deviation of the sampling distribution of sample mean as follows:

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}$

$=\frac{0.33}{\sqrt{40}}\\\\=0.052178\\\\\approx 0.052$

Thus, the standard error of the mean in this experiment is $0.052.

(b)

Compute the probability that the sample mean is between $3.78 and $3.86 as follows:

$P(3.78$

$=P(-0.77$

Thus, the probability that the sample mean is between $3.78 and $3.86 is 0.5587.

(c)

If the difference between the sample mean and the population mean is less than $0.01 then:

$\bar X-\mu_{\bar x}$

Compute the value of $P(\bar X$ as follows:

$P(\bar X$

$=P(Z$

Thus, the probability that the difference between the sample mean and the population mean is less than $0.01 is 0.5754.

(d)

Compute the probability that the sample mean is greater than $3.92 as follows:

$P(\bar X>3.92)=P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}}>\frac{3.92-3.82}{0.052})$

$=P(Z$

Thus, the likelihood that the sample mean is greater than $3.92 is 0.9726.

3 0

3 years ago

3. There are 124 acorns laying on the ground in your yard, it is predicted that the number of acorns will

Gennadij [26K]

Answer:

507,409

Step-by-step explanation:

if every 5 days the number quadruples (x4), and we want to know how many acorns fall after 30 days, we can divide 30 by 5 so we only have to calculate for the amount of time the take to quadruple.

30 ÷ 5 = 6

so we only have to quadruple the acorns 6 times.

if we start off with 124 acorns, this is what it will look like:

Day 0: 124 acorns

Day 5: 124 x 4 = 496 acorns

Day 10: 496 x 4 = 1984 acorns

Day 15: 1984 x 4 = 7936 acorns

etc... until day 30.

Day 30: 507904 acorns

I hope this was helpful :-)

5 0

3 years ago

5^(-x)+7=2x+4 This was on plato

Setler79 [48]

Answer:

Below

I hope its not too complicated

$x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}$

Step-by-step explanation:

$5^{\left(-x\right)}+7=2x+4\\\\\mathrm{Prepare}\:5^{\left(-x\right)}+7=2x+4\:\mathrm{for\:Lambert\:form}:\quad 1=\left(2x-3\right)e^{\ln \left(5\right)x}\\\\\mathrm{Rewrite\:the\:equation\:with\:}\\\left(x-\frac{3}{2}\right)\ln \left(5\right)=u\mathrm{\:and\:}x=\frac{u}{\ln \left(5\right)}+\frac{3}{2}\\\\1=\left(2\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)-3\right)e^{\ln \left(5\right)\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)}$

$Simplify\\\\\mathrm{Rewrite}\:1=\frac{2e^{u+\frac{3}{2}\ln \left(5\right)}u}{\ln \left(5\right)}\:\\\\\mathrm{in\:Lambert\:form}:\quad \frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}$

$\mathrm{Solve\:}\:\frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}:\quad u=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)\\\\\mathrm{Substitute\:back}\:u=\left(x-\frac{3}{2}\right)\ln \left(5\right),\:\mathrm{solve\:for}\:x$

$\mathrm{Solve\:}\:\left(x-\frac{3}{2}\right)\ln \left(5\right)=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right):\\\quad x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}$

3 0

3 years ago

A cylinder has a base area of 64π m2. Its height is equal to twice the radius. Identify the volume of the cylinder to the neares

photoshop1234 [79]

Step-by-step explanation:

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3 0

2 years ago