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1 year ago

WILL. GIVE BRAINLIEST IF RIGHT!!

Mathematics

1 answer:

Natali [406]1 year ago

3 0

<u>Answer:</u>

The class and exam medians are almost the same.

<u>Step-by-step explanation:</u>

The median of a box-and-whisker plot is the line inside the box.

• As we can see, the medians of class and median are very similar, both around 85. This is why the first option is correct.

•The second option is incorrect because the exam median is not much higher than the class median; the difference is very small.

• The third option is incorrect because exam does not have the lowest median. In fact, exam median is slightly higher than class median.

• The fourth option is incorrect because outliers do not affect the median.

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What’s the value of x in this equation???? Thank u!

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Answer:

x = 9

Step-by-step explanation:

The exterior angle of a triangle is equal to the sum of the 2 opposite interior angles.

50° is an exterior angle of the triangle, thus

38 + 4x - 24 = 50 , that is

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3 years ago

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2 years ago

A sociologist was investigating the ages of grandparents of high school students. From a random sample of 10 high school student

umka2103 [35]

Answer:

$(69.8-71.6) -2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= -8.908$

$(69.8-71.6) +2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= 5.308$

So then the confidence interval for the difference of means is given by:

$-8.908 \leq \mu_M -\mu_F \leq 5.308$

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".

$\bar X_{M}= 69.8$ represent the mean for the age of grandmother

$\bar X_{F}= 71.6$ represent the mean for the age of grandfather

$s_{M}= 8.38$ represent the sample deviation for the age of grandmother

$s_{F}= 6.65$ represent the sample deviation for the age of grandfather

$n_M = n_F= 10$

Solution to the problem

For this case the confidence interval is given by:

$(\bar X_{M} -\bar X_F) \pm t_{\alpha/2} \sqrt{\frac{s^2_M}{n_M} +\frac{s^2_F}{n_F}}$

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n_M +n_F-2=10+10-2=18$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,18)".And we see that $t_{\alpha/2}=2.101$