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1 year ago

Lizzy had test scores of: 72, 94, 108, 60What is the RANGE of her test scores?

Mathematics

1 answer:

Misha Larkins [42]1 year ago

5 0

Answer:

Step-by-step explanation:

To find the range, you have to subtract the biggest/largest number by the smallest number:

108 - 60 = 48

The range is 48.

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A grasshopper jumps off a tree stump. The height, in feet, of the grasshopper above the ground after t seconds is modeled by the

Monica [59]

The height, in feet, of the grasshopper above the ground after t seconds is modeled by the function , which is

$h(t) =-t^2+\frac{4}{3}t+ \frac{1}{4}$

When the grasshopper land on the ground,

$h(t)=0$

that is,

$-t^2+\frac{4}{3}t+ \frac{1}{4}=0$

Multiplying whole equation by 12 to get rid of denominators 3 and 4

$-12t^2+16t +3=0$

$(3-2t)(6t+1)=0 \\ 3-2t=0 , 6t+1=0 \\ t= \frac{3}{2} , \frac{-1}{6}$

And time cant be negative, so the correct option is

$t = \frac{3}{2}=1.5 seconds$

Correct option is C .

0 0

2 years ago

Line AB has an equation of a line y = 5x − 2. Which of the following could be an equation for a line that is parallel to line AB

SpyIntel [72]

Determine the slope of line AB
m = 5

Determine the slope of the lines from the options
First option: y = 5x + 3, the slope is 5
Second option: y = (1/5)x + 3, the slope is 1/5
Third option: y = -5x + 3, the slope is -5
Fourth option: y = (-1/5)x + 3, the slope is -1/5

Parallel lines are similar in the slope. So the line which is parallel to line AB must have the slope of 5.

The answer is first option.

4 0

2 years ago

A team averaging 110 points is likely to do very well during the regular season. The coach of your team has hypothesized that yo

aev [14]

Answer:

1. M=108

2. μ=110

3. In the explanation.

4. Test statistic t = -1.05

5. P-value = 0.1597

Step-by-step explanation:

The question is incomplete: to solve this problem, we need the sample information: size, mean and standard deviation.

We will assume a sample size of 10 matches, a sample mean of 108 points and a sample standard deviation of 6 points.

1. The mean points is the sample points and has a value of 108 points.

2. The null hypothesis is H0: μ=110, meaning that the mean score is not significantly less from 110 points.

3. This is a hypothesis test for the population mean.

The claim is that the mean score is significantly less than 110.

Then, the null and alternative hypothesis are:

$H_0: \mu=110\\\\H_a:\mu< 110$

The significance level is 0.05.

The sample has a size n=10.

The sample mean is M=108.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=6.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{6}{\sqrt{10}}=1.9$

4. Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{108-110}{1.9}=\dfrac{-2}{1.9}=-1.05$

The degrees of freedom for this sample size are:

$df=n-1=10-1=9$

5. This test is a left-tailed test, with 9 degrees of freedom and t=-1.05, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=P(t$

As the P-value (0.1597) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the mean score is significantly less than 110.

7 0

2 years ago

5. Mrs. Jones is handing out candy to trick-or-

mrs_skeptik [129]

ANSWER. BBB

Step-by-step explanation:

6 0

2 years ago

N 2004, the General Social Survey (which uses a method similar to simple random sampling) asked, "Do you consider yourself athle

zmey [24]

<u>Answer-</u>

The standard error of the confidence interval is 0.63%

<u>Solution-</u>

Given,

n = 2373 (sample size)

x = 255 (number of people who bought)

The mean of the sample M will be,

$M=\frac{x}{n} =\frac{255}{2373} =0.1075$

Then the standard error SE will be,

$SE=\sqrt{\frac{M\times (1-M)}{n}}$

$SE=\sqrt{\frac{0.1075\times (1-0.1075)}{2373}}=\sqrt{\frac{0.0959}{2373}}=0.0063=0.63\%$

Therefore, the standard error of the confidence interval is 0.63%

6 0

2 years ago