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2 years ago

I need to know what goes in boxes 10 points :)

Mathematics

2 answers:

Evgen [1.6K]2 years ago

7 0

Answer:

1/10 + 1/2

= 1/10 + 5/10

= 6/10 or 3/5

vredina [299]2 years ago

7 0

Answer:

6/10

Step-by-step explanation:

$\frac{1}{10} + \frac{1}{2} = \frac{1}{10} + \frac{5}{10} = \frac{6}{10}$

Further:

6/10 can be reduced to 3/5. The question does not ask for it, so you can ignore it.

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Write an equation in point-slope form of the line having the given slope that contains the given point. m = -2, (4, -2)write an

alex41 [277]

Answer:

Point-slope form: y+2=-2(x-4)

6 0

2 years ago

A sample of size 6 will be drawn from a normal population with mean 61 and standard deviation 14. Use the TI-84 Plus calculator.

AVprozaik [17]

Answer:

$X \sim N(61,14)$

Where $\mu=61$ and $\sigma=14$

Since the distribution for X is normal then the distribution for the sample mean is also normal and given by:

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

$\mu_{\bar X} = 61$

$\sigma_{\bar X}= \frac{14}{\sqrt{6}}= 5.715$

So then is appropiate use the normal distribution to find the probabilities for $\bar X$

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". The letter $\phi(b)$ is used to denote the cumulative area for a b quantile on the normal standard distribution, or in other words: $\phi(b)=P(z$

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:

$X \sim N(61,14)$

Where $\mu=61$ and $\sigma=14$

Since the distribution for X is normal then the distribution for the sample mean $\bar X$ is also normal and given by:

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

$\mu_{\bar X} = 61$

$\sigma_{\bar X}= \frac{14}{\sqrt{6}}= 5.715$

So then is appropiate use the normal distribution to find the probabilities for $\bar X$

8 0

3 years ago

Math scores on the SAT exam are normally distributed with a mean of 514 and a standard deviation of 118. If a recent test-taker

LuckyWell [14K]

Answer:

Probability that the student scored between 455 and 573 on the exam is 0.38292.

Step-by-step explanation:

We are given that Math scores on the SAT exam are normally distributed with a mean of 514 and a standard deviation of 118.

Let X = Math scores on the SAT exam

So, X ~ Normal( $\mu=514,\sigma^{2} =118^{2}$ )

The z score probability distribution for normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean score = 514

$\sigma$ = standard deviation = 118

Now, the probability that the student scored between 455 and 573 on the exam is given by = P(455 < X < 573)

P(455 < X < 573) = P(X < 573) - P(X $\leq$ 455)

P(X < 573) = P( $\frac{X-\mu}{\sigma}$ < $\frac{573-514}{118}$ ) = P(Z < 0.50) = 0.69146

P(X $\leq$ 2.9) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{455-514}{118}$ ) = P(Z $\leq$ -0.50) = 1 - P(Z < 0.50)

= 1 - 0.69146 = 0.30854

The above probability is calculated by looking at the value of x = 0.50 in the z table which has an area of 0.69146.

Therefore, P(455 < X < 573) = 0.69146 - 0.30854 = 0.38292

Hence, probability that the student scored between 455 and 573 on the exam is 0.38292.

7 0

3 years ago

Create an equation with at least two grouping symbols for which there is no solution

dedylja [7]

5+7-4=6
6-4=2
I’ve created two equations with no solution

5 0

2 years ago

The enrollment at East Valley High School over a six-year period is displayed in the scatterplot.

Anna [14]

Answer:

we need a scatterplot

Step-by-step explanation:

If you provided one id be happy to help :)

7 0

2 years ago