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3 years ago

How do you solve this problem 50=t/10

Mathematics

2 answers:

Masja [62]3 years ago

8 0

Answer:

500

Step-by-step explanation:

50=t/10

Move the equation over the equal sign which means you change the operation from division to multiplication

50*10 = 500

Delicious77 [7]3 years ago

7 0

Answer:

Step-by-step explanation:

$50=\dfrac{t}{10}\qquad\text{multiply both sides by 10}\\\\(50)(10)=\left(\dfrac{t}{10}\right)(1)\\\\500=\dfrac{10t}{10}\qquad\text{cancel 10}\\\\500=t\to t=500$

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Can anyone please help me this? stuck with this problem for hours

Kipish [7]

Answer:

P=0.61 or 61%

Step-by-step explanation:

If we pick a random point within the circle, it can fall in the white area or in the pink area. Since the sum of both areas makes the sample space or the total area of the circle, we can compute the probabilities according to the ratio of the areas with respect to the total area, that is:

$\text {Probability of falling in the white area} = \frac{\text{White area}}{\text{Circle area}}$

The area of the circle is

$A_c=\pi r^2$

The area of both the triangles is

$A_t=2*\frac{bh}{2}$

Where r=4 cm, b=3 cm, h= (4+2.5) cm = 6.5 cm

Then we have:

$A_c=\pi 4^2=16\pi \ cm^2=50.27\ cm^2$

$A_t=19.5\ cm^2$

The white area is obtained by subtracting both areas

$A_w=A_c-A_t=50.27-19.5=30.77\ cm^2$

So the probability is

$P=\frac{30.77}{50.27}=0.61$

5 0

4 years ago

Suppose a simple random sample of size n is drawn from a large population with mean mu and standard deviation sigma. The samplin

Sidana [21]

Answer:

$X \sim N(\mu,\sigma)$

Where $\mu$ the mean and $\sigma$ the deviation

Since the distribution for X is normal then we can conclude that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

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

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

Step-by-step explanation:

Previous concepts

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".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:

$X \sim N(\mu,\sigma)$

Where $\mu$ the mean and $\sigma$ the deviation

Since the distribution for X is normal then we can conclude that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

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

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

6 0

3 years ago

Read 2 more answers

Can someone help me out with this?