All categories

2 years ago

How to write 0.12 repeating as a fractiom

Mathematics

1 answer:

frosja888 [35]2 years ago

5 0

Answer:

4/33

Step-by-step explanation:

do the normal steps to conversion

You might be interested in

A firm offers routine physical examinations as part of a health service program for its employees. The exams showed that 8% of t

Alexxandr [17]

Answer: A. 0.20

Step-by-step explanation:

Let A be the event of employees needed corrective shoes and B be the event that they needed major dental work .

We are given that : $P(A)=0.08\ ;\ P(B)=0.15\ ;\ P(A\cap B)=0.03$

We know that $P(A\cup B)=P(A)+P(B)-P(A\cap B)$

Then, $P(A\cup B)=0.08+0.15-0.03= 0.20$

Hence, the probability that an employee selected at random will need either corrective shoes or major dental work : $P(A\cup B)= 0.20$

hence, the correct option is (A).

6 0

3 years ago

Find an exponential function that fits the information (hint: putting them in a table may help you see).

Vilka [71]

Answer:

f(x)=8*(3.5)^x

Step-by-step explanation:

2=98

3= 98r

4= 98r^2

5= 98r^3

6= 98r^4

98r^4/98, 98 crosses out and its just r^4. r^4= 14706.125/98

r=3.5

So you plug that in and you get f(x)=8*(3.5)^x

I hope this helped. XD

7 0

2 years ago

If f(x) = x* and g(x) = 3+8x?, find g(f(x)).

zhuklara [117]

g(8x-3)=x

have a good day

6 0

2 years ago

A dilation by a factor of 7/8 maps triangle ABC to triangle XYZ. Segment BC has a length of 4 cm.

anastassius [24]

I believe the answer should be 3 and 1/2 which is 3.5 because I did the factor 7/8 times 4 and I got 3 and 1/2 I’m not sure

5 0

2 years ago

Suppose that you had the following data set. 500 200 250 275 300 Suppose that the value 500 was a typo, and it was suppose to be

hodyreva [135]

Answer:

$\bar X_B = \frac{\sum_{i=1}^5 X_i}{5} =\frac{500+200+250+275+300}{5}=\frac{1525}{5}=305$

$s_B = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(500-305)^2 +(200-305)^2 +(250-305)^2 +(275-305)^2 +(300-305)^2)}{5-1}} = 115.108$

$\bar X_A = \frac{\sum_{i=1}^5 X_i}{5} =\frac{-500+200+250+275+300}{5}=\frac{525}{5}=105$

$s_A = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(-500-105)^2 +(200-105)^2 +(250-105)^2 +(275-105)^2 +(300-105)^2)}{5-1}} = 340.221$

The absolute difference is:

$Abs = |340.221-115.108|= 225.113$

If we find the % of change respect the before case we have this:

$\% Change = \frac{|340.221-115.108|}{115.108} *100 = 195.57\%$

So then is a big change.

Step-by-step explanation:

The subindex B is for the before case and the subindex A is for the after case

Before case (with 500)

For this case we have the following dataset:

500 200 250 275 300

We can calculate the mean with the following formula:

$\bar X_B = \frac{\sum_{i=1}^5 X_i}{5} =\frac{500+200+250+275+300}{5}=\frac{1525}{5}=305$

And the sample deviation with the following formula:

$s_B = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(500-305)^2 +(200-305)^2 +(250-305)^2 +(275-305)^2 +(300-305)^2)}{5-1}} = 115.108$

After case (With -500 instead of 500)

For this case we have the following dataset:

-500 200 250 275 300

We can calculate the mean with the following formula:

$\bar X_A = \frac{\sum_{i=1}^5 X_i}{5} =\frac{-500+200+250+275+300}{5}=\frac{525}{5}=105$

And the sample deviation with the following formula:

$s_A = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(-500-105)^2 +(200-105)^2 +(250-105)^2 +(275-105)^2 +(300-105)^2)}{5-1}} = 340.221$

And as we can see we have a significant change between the two values for the two cases.

The absolute difference is:

$Abs = |340.221-115.108|= 225.113$

If we find the % of change respect the before case we have this:

$\% Change = \frac{|340.221-115.108|}{115.108} *100 = 195.57\%$

So then is a big change.

8 0

3 years ago