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3 years ago

Write the following tepeating decmial 0.35 as a fraction

Mathematics

2 answers:

vichka [17]3 years ago

6 0

I believe it is 35 over 99 sorry if it’s wrong <3

Mila [183]3 years ago

4 0

Answer:

0.35 is equal to 7/20

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Number 5 am stuck pls help

Norma-Jean [14]

Answer:

im going to be honest i havent done this since las year and so i dont remember the value of that line outside but i think its 70*

Step-by-step explanation:

reason is because the c is originally 180* but it has more of and angle so the angles obv increases and you can get the angle by multiplying 70*times 7x and divide by 3

i say 3 because of the amount of angles in a triangle and the line outside if i remember correctly is the "angles" of a triangle and that would be your answer

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3 years ago

What is 7/21 in its simplest form

olasank [31]

The answer is 1/3 because 7 can go into 21 so then 7 goes into 7 one time and then it goes into 21 three times so then you would get 1/3

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4 years ago

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What is the product (-2a^2+s)(5a^2-6s)

Mariulka [41]

The answer that you should get 35as^2-30s

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3 years ago

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A researcher compares the effectiveness of two different instructional methods for teaching electronics. A sample of 138 student

yan [13]

Answer:

The 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 is (-8.04, 0.84).

Step-by-step explanation:

The (1 - <em>α</em>)% confidence interval for the difference between population means is:

$CI=(\bar x_{1}-\bar x_{2})\pm z_{\alpha/2}\times \sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}$

The information provided is as follows:

$n_{1}= 138\\n_{2}=156\\\bar x_{1}=61\\\bar x_{2}=64.6\\\sigma_{1}=18.53\\\sigma_{2}=13.43$

The critical value of <em>z</em> for 98% confidence level is,

$z_{\alpha/2}=z_{0.02/2}=2.326$

Compute the 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 as follows:

$CI=(\bar x_{1}-\bar x_{2})\pm z_{\alpha/2}\times \sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}$

$=(61-64.6)\pm 2.326\times\sqrt{\frac{(18.53)^{2}}{138}+\frac{(13.43)^{2}}{156}}\\\\=-3.6\pm 4.4404\\\\=(-8.0404, 0.8404)\\\\\approx (-8.04, 0.84)$

Thus, the 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 is (-8.04, 0.84).

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3 years ago

What is the solution to the equation 7,- 5* ++5

maw [93]

Answer:

the answer to the problem is h=7

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3 years ago