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4 years ago

Which is an example of discrete data?

Mathematics

2 answers:

son4ous [18]4 years ago

5 0

Answer:

The answer is C. The number of shelves in a bookcase

Step-by-step explanation:

The answer is C. number of shelves in a bookcase because discrete data can be counted using whole numbers. Height, width, and weight are continuous concepts that are measured in units and fractions of units.

tatiyna4 years ago

4 0

Answer:

Step-by-step explanation:

You might be interested in

Solve for X. 6x – 9 = -27

MissTica

Answer:

x = -3

Step-by-step explanation:

6x - 9 = -27 (Given)

6x = -18 (Add 9 on both sides.)

x = -3 (Divide 6 on both sides.)

6 0

3 years ago

Read 2 more answers

Leo sold candy bars for a fundraiser. He spent $25 on a box of candy bars and sold them for $2 each. His profit (after earning b

kykrilka [37]

As given by the question

There are given that the $2c-$25=$69.

Now,

$\begin{gathered} $2c-$25=$69$ \\ $2c-$25+25=$69$+25 \\ 2c=$69$+25 \\ 2c=94 \\ c=\frac{94}{2} \\ c=47\text{ bars} \end{gathered}$

Hence, the correct answer is C.

4 0

1 year ago

For the hypothesis test H0: μ = 10 against H1: μ >10 and variance known, calculate the Pvalue for each of the following test

Basile [38]

Answer:

a) $p_v =P(Z>2.05)=1-P(z$

b) $p_v =P(Z>-1.84)=1-P(z$

c) $p_v =P(Z>0.4)=1-P(z$

Step-by-step explanation:

Some previous concepts

The p-value is the probability of obtaining the observed results of a test, assuming that the null hypothesis is correct.

A z-test for one mean "is a hypothesis test that attempts to make a claim about the population mean(μ)".

The null hypothesis attempts "to show that no variation exists between variables or that a single variable is no different than its mean"

The alternative hypothesis "is the hypothesis used in hypothesis testing that is contrary to the null hypothesis"

Hypothesis

Null hypothesis: $\mu=10$

Alternative hypothesis: $\mu >10$

If the random variable is distributed like this: $X \sim N(\mu,\sigma)$

We assume that the variance is known so the correct test to apply here is the z test to compare means, the statistic is given by the following formula:

$z_o=\frac{\bar X -\mu}{\sigma}$

Since we have the values for the statistic already calculated we can calculate the p value using the following formulas:

Part a

$p_v =P(Z>2.05)=1-P(z$

And in order to find the answer using excel we can use the following code:

"=1-NORM.DIST(2.05,0,1,TRUE)"

Part b

$p_v =P(Z>-1.84)=1-P(z$

And in order to find the answer using excel we can use the following code:

"=1-NORM.DIST(-1.84,0,1,TRUE)"

Part c

$p_v =P(Z>0.4)=1-P(z$

And in order to find the answer using excel we can use the following code:

"=1-NORM.DIST(0.4,0,1,TRUE)"

Conclusions

If we use a reference value for the significance, let's say $\alpha=0.05$ . For part a the $p_v$ so then we can reject the null hypothesis at this significance level.