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2 years ago

Can someone please help me with this question?

Mathematics

2 answers:

Anna007 [38]2 years ago

8 0

$\qquad\qquad\huge\underline{{\sf Answer}}$

As per the given graph we can observe the number of students which watched TV for certain intervals.

According to question, we have to find the common interval of minutes students spent watching TV~

That means : we have to find the time interval in which most of the students prefer to watch TV ~

Hence, the most common interval is :

$\qquad \tt \dashrightarrow \:150 - 179 \: \: \: minutes$

MrRissso [65]2 years ago

4 0

Answer:

150-179 minutes

Step-by-step explanation:

The most common interval (the tallest graph) of minutes students spent watching watching TV?

The correct answer should be 150-179 minutes because it's the most common, meaning that a lot of students spent 150-179 minutes watching TV.

Hope this helps!

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Which graph shows the solution to the inequality -3x-7x<20

hjlf

A graph of this inequality would be a number line, with the number -2 circled (not filled in), and the number line to the right of -2 shaded.

To solve this, you will combine like terms first:

-3x-7x < 20
-10x < 20

Divide both sides by -10:
-10x/-10 < 20/-10

When you divide an inequality by a negative number, you flip the inequality symbol:

x > -2

5 0

3 years ago

The Rocky Mountain district sales manager of Rath Publishing Inc., a college textbook publishing company, claims that the sales

mihalych1998 [28]

Answer:

We conclude that the mean number of calls per salesperson per week is more than 37.

Step-by-step explanation:

We are given that the Rocky Mountain district sales manager of Rath Publishing Inc., a college textbook publishing company, claims that the sales representatives make an average of 37 sales calls per week on professors.

To investigate, a random sample of 41 sales representatives reveals that the mean number of calls made last week was 40. The standard deviation of the sample is 5.6 calls.

Let $\mu$ = true mean number of calls per salesperson per week.

SO, Null Hypothesis, $H_0$ : $\mu \leq$ 37 {means that the mean number of calls per salesperson per week is less than or equal to 37}

Alternate Hypothesis, $H_A$ : $\mu$ > 37 {means that the mean number of calls per salesperson per week is more than 37}

The test statistics that will be used here is One-sample t test statistics as we don't know about the population standard deviation;

T.S. = $\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean number of calls made last week = 40

s = sample standard deviation = 5.6 calls

n = sample of sale representatives = 41

So, test statistics = $\frac{40-37}{\frac{5.6}{\sqrt{41} } }$ ~ $t_4_0$

= 3.43

Hence, the value of test statistics is 3.43.

Now at 0.025 significance level, the t table gives critical value of 2.021 at 40 degree of freedom for right-tailed test. Since our test statistics is more than the critical value of t as 3.43 > 2.021, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the mean number of calls per salesperson per week is more than 37.

4 0

3 years ago

120 is what percent of 90? Lo in

allsm [11]

Answer:

133.33%

Step-by-step explanation:

120= 120/90

the word OF = multiplication

120/90 x 100/100=133.33/100

133.33/100=133.33%