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Nadusha1986 [10]
3 years ago
13

Cole and his friends made a line plot of hours they spend watching TV after school, which is the difference between the greatest

amount of time spent watching TV at the least amount time.
Mathematics
1 answer:
mario62 [17]3 years ago
5 0

There are no numbers soo…..

The answer is KEEP WATCHING TV!!!

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Write the equation of the line graphed below in slope-intercept form (y=mx+b)?​
ioda
Y = -1/2x - 1

Tell me if you need explanation
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3 years ago
every tenth visitor to an ice cream shop gets free sprinkler. every twelvth visitor gets free hot fudge. which visitor each day
DENIUS [597]


person number 60

10 20 30 40 50<u> 60</u>

12 14 36 48<u> 60</u>

It id in both of their times tables.<u />

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PLEEASe help me on this table question!!! it's easy
zhuklara [117]
I'm pretty sure that it's the first one: direct variation; y=4x.
8 0
3 years ago
If you are solving y 2+2y=48 by completing the square, the next line would be
snow_tiger [21]

For this case we have the following equation:

y ^ 2 + 2y = 48

By completing squares we have:

Step 1:

Add (\frac {b} {2}) ^ 2 on both sides:

y ^ 2 + 2y + 1 = 48 + 1

Step 2:

We rewrite the equation:

(y + 1) ^ 2 = 49

Step 3:

We solve the equation:

y = \pm \sqrt {49} -1\\y = \pm7-1

Solutions:

y = 6\\y = -8

Answer:

the next line would be:

y ^ 2 + 2y + 1 = 48 + 1


6 0
3 years ago
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A manager of a grocery store wants to determine if consumers are spending more than the national average. The national average i
strojnjashka [21]

The valid conclusions for the manager based on the considered test is given by: Option

<h3>When do we perform one sample z-test?</h3>

One sample z-test is performed if the sample size is large enough (n  > 30) and we want to know if the sample comes from the specific population.

For this case, we're specified that:

  • Population mean = \mu = $150
  • Population standard deviation = \sigma = $30.20
  • Sample mean = \overline{x} = $160
  • Sample size = n = 40 > 30
  • Level of significance = \alpha = 2.5% = 0.025
  • We want to determine if the average customer spends more in his store than the national average.

Forming hypotheses:

  • Null Hypothesis: Nullifies what we're trying to determine. Assumes that the average customer doesn't spend more in the store than the national average. Symbolically, we get: H_0: \mu_0 \leq \mu = 150
  • Alternate hypothesis: Assumes that customer spends more in his store than the national average. Symbolically H_1: \mu_0 > \mu = 150

where \mu_0 is the hypothesized population mean of the money his customer spends in his store.

The z-test statistic we get is:

z = \dfrac{\overline{x} - \mu_0}{\sigma/\sqrt{n}} = \dfrac{160 - 150}{30.20/\sqrt{40}} \approx 2.094

The test is single tailed, (right tailed).

The critical value of z at level of significance 0.025 is 1.96

Since we've got 2.904 > 1.96, so we reject the null hypothesis.

(as for right tailed test, we reject null hypothesis if the test statistic is > critical value).

Thus, we accept the alternate hypothesis that customer spends more in his store than the national average.

Learn more about one-sample z-test here:

brainly.com/question/21477856

3 0
2 years ago
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