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3 years ago

Plz help me I’m struggling

Mathematics

1 answer:

Delicious77 [7]3 years ago

3 0

Answer:

i can help if you zoom in and repost it

Step-by-step explanation:

You might be interested in

The producer of the news station posted an article about the high school's football championship ceremony on anew website. The w

sasho [114]

Incomplete question. The full question read;

The producer of the news station posted an article about the high school’s football championship ceremony on a new website. The website had 500 views after four hours. Create a table to show how many views the website would have had after the first, second, and third hours after posting if the website receives views at the same rate. How many views would the website receive after 5 hours?

Answer:

625 views

Step-by-step explanation:

Remember, we are told to base our calculation on the assumption that the website receives views at the same rate.

Hence, if after 4 hours there were 500 views, it means the average views per hour would be 500/4 = 125 views. So for every hour, there would be 125 added views to the total.

In other words,

First hour: 125 views

Second hour: 250 views (125 views + 125 views)

Third hour: 375 views (250 views + 125 views)

Fourth hour: 500 views (375 views + 125 views)

Fifth hour: 625 views (500 views + 125 views)

7 0

2 years ago

What is the Midpoint of CD given the coordinates C(-4,8) and D(8,-4)??

alina1380 [7]

Use this formula...
(-4 + 8) / 2 , (8 + (-4)) / 2
From there you get...
4 / 2 , 4 / 2
Simplify and you get
(2 , 2)

3 0

3 years ago

Read 2 more answers

A manufacturer of running shoes knows that the average lifetime for a particular model of shoes is 15 months. Someone in the res

AveGali [126]

Answer:

The decision rule is

Fail to reject the null hypothesis

The conclusion is

There is no sufficient evidence to show that the designer's claim of a better shoe is supported by the trial results.

Step-by-step explanation:

From the question we are told that

The population mean is $\mu = 15 \ months$

The sample size is n = 25

The sample mean is $\= x = 17 \ months$

The standard deviation is $s = 5.5 \ months$

Let assume the level of significance of this test is $\alpha = 0.05$

The null hypothesis is $H_o : \mu = 15$

The alternative hypothesis is $H_a : \mu > 17$

Generally the degree of freedom is mathematically represented as

$df = n -1$

=> $df = 25 -1$

=> $df = 24$

Generally the test statistics is mathematically represented as

$t = \frac{\= x- \mu }{\frac{s}{\sqrt{n} } }$

=> $t = \frac{ 17 - 15 }{\frac{5.5}{\sqrt{25} } }$

=> $t = 1.8182$

Generally from the student t distribution table the probability of obtaining $t = 1.8182$ to the right of the curve at a degree of freedom of $df = 24$ is

$p-value = P(t > 0.18182 ) = 0.4286$