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4 years ago

Oof

Mathematics

2 answers:

Marizza181 [45]4 years ago

8 0

Answer:

triangle C has the largest area.

Step-by-step explanation:

From the the three triangles A, B, C based on the area occupied by the triangles, it can be seen that triangle C occupies the largest area.

On the centimetre grid, let each box represent one unit. For triangle A, the base is 2 boxes (i.e 2 units) and the height is 3 units, For triangle B, the base is 2 units and the height is 3 units while for triangle C, the base is 3 units and the height is 3 unit. The area of a triangle is given by:

Area = $\frac{1}{2}base*height$

For triangle A:

Area = $\frac{1}{2}base*height$ = $\frac{1}{2} *2*3=3 \ unit^2$

For triangle B:

Area = $\frac{1}{2}base*height$ = $\frac{1}{2} *2*3=3 \ unit^2$

For triangle C:

Area = $\frac{1}{2}base*height$ = $\frac{1}{2} *3*3=4.5 \ unit^2$

Therefore, triangle C has the largest area.

Elina [12.6K]4 years ago

5 0

Answer:

triangle C

Step-by-step explanation:

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13. Interpret the phrase ""margin of sampling error is ±1 percentage point.""

Ulleksa [173]

Answer:

On this case a margin of error of $\pm 1 \%$ means that the true population proportion is 1% above or below the estimated proportion calculated from the sample. And this value helps in order to find the limits for a confidence interval with a confidence given.

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".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Solution to the problem

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

4 0

3 years ago

. The time required for a technician to machine a specific component is normally distributed with a mean of 2 hours and a standa

erma4kov [3.2K]

Answer:

a) There is a 3.84% probability that the technician can machine one component in 1.5 hours or less.

b) There is a 0.42% probability that the technician will require at least 2.75 hours to complete one component

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by

$Z = \frac{X - \mu}{\sigma}$

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X.

In this problem, we have to be careful. The mean is in hours, while the standard deviation is in minutes. I am going to work with both in hours, as the problem states. 17 minutes is 0.283 hours, so:

$\mu = 2, \sigma = 0.283$

(a.) What is the probability that the technician can machine one component in 1.5 hours or less?

This probability is the pvalue of the Zscore when $X = 1.5$ . So:

$Z = \frac{1.5 - 2}{0.283}$

$Z = \frac{-0.5}{0.283}$

$Z = -1.77$

$Z = -1.77$ has a pvalue of 0.0384.

This means that there is a 3.84% probability that the technician can machine one component in 1.5 hours or less.

(b.) What is the probability that the technician will require at least 2.75 hours to complete one component?

The pvalue of the score of $X = 2.75$ is the probability that the technican will require less than 2.75 hours to complete one component. The probability that he will require at least 2.75 hours to complete one component is 1 subtracted by this pvalue. So: