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insens350 [35]
2 years ago
5

Joseph is going to paint the panels of a fence. He is going to use the same repeating pattern for the entire fence. The rule is

"orange, yellow, pink, green, purple." If there are 98 panels in the fence, how many will be painted green? Enter your answer in the box.
Mathematics
1 answer:
Marianna [84]2 years ago
8 0

19 times Answer:

Step-by-step explanation:

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4252.50=(18000)(r)(54)
drek231 [11]

Step-by-step explanation:

do you want to find r?

so do 4252.50/ (18000*54)

5 0
3 years ago
Last nights travel programme had viewing figures of 3,429,111. What is the place value of the 4
Phantasy [73]
The 4 is in the hundred thousands place hope i helped
7 0
3 years ago
Read 2 more answers
Urgently needed<br> see image
Lorico [155]

y2-y1 =M(x2-x1)

Ok Sir, you gave points : (2,1) and (3,4)

4-1/3-2 = 3

Ok we know our slope is 3, now pick any of the two points, and make an equation for this line, so lets go ahead and pick #1, (2,1)

Formula is same as before

y-1=3x-6

y=3x-5

I think its correct, pick as brainless sir, thanks.

7 0
2 years ago
If you are adding two fractions that are both greater than 1/2, what must be true about the sum? Give three examples to support
anastassius [24]

Answer:

Step-by-step

The sum must be greater than 1 because

1/2 + 1/2= 2/2 or 1

ex

3/4 + 8/4= 11/4

9/16+15/16=24/16

8/8+9/8=17/8

5 0
3 years ago
A certain pen has been designed so that true average writing lifetime under controlled conditions (involving the use of a writin
Morgarella [4.7K]

Answer:

(a) We reject our null hypothesis.

(b) We fail to reject our null hypothesis.

(c) We fail to reject our null hypothesis.

Step-by-step explanation:

We are given that a certain pen has been designed so that true average writing lifetime under controlled conditions (involving the use of a writing machine) is at least 10 hr.

A random sample of 18 pens is selected.

<u><em>Let </em></u>\mu<u><em> = true average writing lifetime under controlled conditions</em></u>

So, Null Hypothesis, H_0 : \mu \geq 10 hr   {means that the true average writing lifetime under controlled conditions is at least 10 hr}

Alternate Hypothesis, H_A : \mu < 10 hr    {means that the true average writing lifetime under controlled conditions is less than 10 hr}

<u>The test statistics that is used here is one-sample t test statistics;</u>

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

where, \bar X = sample mean

             s = sample standard deviation

             n = sample size of pens = 18

          n - 1 = degree of freedom = 18 -1 = 17

<u>Now, the decision rule based on the critical value of t is given by;</u>

  • If the value of test statistics is more than the critical value of t at 17 degree of freedom for left-tailed test, then <u>we will not reject our null hypothesis</u> as it will not fall in the rejection region.
  • If the value of test statistics is less than the critical value of t at 17 degree of freedom for left-tailed test, then <u>we will reject our null hypothesis</u> as it will fall in the rejection region.

(a) Here, test statistics, t = -2.4 and level of significance is 0.05.

<em>Now, at 0.05 significance level, the t table gives critical value of -1.74 at 17 degree of freedom.</em>

Here, clearly the value of test statistics is less than the critical value of t as -2.4 < -1.74, so we reject our null hypothesis.

(b) Here, test statistics, t = -1.83 and level of significance is 0.01.

<em>Now, at 0.051 significance level, the t table gives critical value of -2.567 at 17 degree of freedom.</em>

Here, clearly the value of test statistics is more than the critical value of t as -2.567 < -1.83, so we fail to reject our null hypothesis.

(c) Here, test statistics, t = 0.57 and level of significance is not given so we assume it to be 0.05.

<em>Now, at 0.05 significance level, the t table gives critical value of -1.74 at 17 degree of freedom.</em>

Here, clearly the value of test statistics is more than the critical value of t as  -1.74 < 0.57, so we fail to reject our null hypothesis.

8 0
2 years ago
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