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zhenek [66]
3 years ago
11

What is 1/8 / 3/4 I am so confused​

Mathematics
2 answers:
faust18 [17]3 years ago
7 0
1/8 / 3/4 it is 0.1666
lidiya [134]3 years ago
6 0

Answer:

\frac{1}{6}

Step-by-step explanation:

1) \frac{1}{8}÷\frac{3}{4}

2) Switch the numbers in the second fraction so that it looks like this:

         \frac{1}{8}÷\frac{4}{3}

3) Now reduce by 2 and solve.

 \frac{1}{8}·\frac{4}{3} =

 \frac{1}{2}·\frac{1}{3} = \frac{1}{6}

i hope this helps.. :)

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There are 200 balls of the same size inside a dark room. There are 100 white balls, 55 red balls, and 45 black balls. If one stu
Komok [63]

Answer:

Probability of picking Red ball is 27.5%

Step-by-step explanation:

Given:

Total Number of balls = 200

Number of White balls = 100

Number of Red balls = 55

Number of Black balls = 45

We need to find the probability of the ball picked to be red.

Solution:

Now we know that;

Probability is equal to number of Possible Outcomes divide by total number of Outcomes multiplied by 100.

framing in equation form we get;

P(Red) = \frac{\textrm{Number of Red balls}}{\textrm{Total Number of balls}} \times 100

P(Red) = \frac{55}{200}\times100

P(Red) = 27.5\%

Hence Probability of picking Red ball is 27.5%.

6 0
3 years ago
The head librarian at the Library of Congress has asked her assistant for an interval estimate of the mean number of books check
Verdich [7]

Answer:

n=(\frac{1.960(150)}{90})^2 =10.67 \approx 11

So the answer for this case would be n=11 rounded up to the nearest integer

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X represent the sample mean for the sample  

\mu population mean (variable of interest)

\sigma =150 represent the population standard deviation

n represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

The confidence interval for this case is given by: (740, 920)

We can find the estimate for the mean and we got:

\bar X = \frac{740+920}{2} = 830

and the margin of error is given by :

ME = \frac{920-740}{2}= 90

The margin of error is given by this formula:

ME=z_{\alpha/2}\frac{s}{\sqrt{n}}    (a)

And on this case we have that ME =90 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

n=(\frac{z_{\alpha/2} s}{ME})^2   (b)

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025;0;1)", and we got z_{\alpha/2}=1.960, replacing into formula (b) we got:

n=(\frac{1.960(150)}{90})^2 =10.67 \approx 11

So the answer for this case would be n=11 rounded up to the nearest integer

5 0
3 years ago
A student likes to use the substitution method for systems of equation. How can he use it with a system that is not in the prope
Kay [80]

Answer:

Step-by-step explanation:

We have given:

-2x+y=4     ---------equation1

3x+4y=49 ---------equation 2

We will solve the 1st equation for y and substitute the value into the 2nd equation.

-2x+y=4     ---------equation1

Move the values to the R.H.S except y

y = 2x+4

Now substitute the value of y in  2nd equation:

3x+4y=49

3x+4(2x+4)=49

3x+8x+16=49

Combine the like terms:

3x+8x=49-16

11x=33

Now divide both the sides by 11

11x/11 = 33/11

x= 3

Now substitute the value of x in any of the above equations: We will substitute the value in equation 1:

-2x+y=4

-2(3)+y=4

-6+y=4

Combine the constants:

y=4+6

y = 10

Thus the solution set of (x,y) is {(3,10)}....

8 0
3 years ago
The repair costs for five cars which were crashed by a safety testing organization were as follows: $100, $120, $180, $220, and
ruslelena [56]
The mean is $150, I believe.
7 0
3 years ago
Read 2 more answers
Please answer this quickly ​
Brums [2.3K]

Answer:

Step-by-step explanation:

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3 years ago
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