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3 years ago

Answer along with how you got the answer 3/8 + 1/8 + 2/8 = ?

Mathematics

1 answer:

zzz [600]3 years ago

3 0

Answer:

6/8 simplified form 3/4

Step-by-step explanation:

Given: 3/8 + 1/8 + 2/8

Here we have to find the sum of the fractions.

All the fractions are like fractions.

So we can add the numerators and keep the denominator as it is. (We should not add the denominator)

Therefore,

3/8 + 1/8 + 2/8 = (3 + 1 + 2)/8

= 6/8

We can simplify this fraction by dividing both the numerator and the denominator by 2.

= 6/8

= 3/4

Hope this is will help you to understand the concept.

Thank you.

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How to solve it to put 65% of his Christmas money in the bank.If he had $70 left after he put the money in the bank. How much mo

kow [346]

Answer:

$200

Step-by-step explanation:

Let X be the amount of Christmas money he had at the beginning.

He put 65% of X in the bank. This is 0.65x

He has $70 left which is equal to 0.35x

We need to write an equation to find the value of x.

70 = 0.35x

Divide by 0.35 to find the value of x.

70/0.35 = 0.35x/0.35

200 = x

He received $200 for Christmas.

8 0

3 years ago

4n^(2)+4n=3 Please help with the work

Kaylis [27]

Answer:

n= 1/2 ( $\frac{1}{2}$ ) or (0.5)

AND

n= -3/2 ( $-\frac{3}{2}$ ) or (-1.5)

Step-by-step explanation:

4n^2+4n=3

step1: move 3 to the other side and make the equation equal to zero.

4n^2+4n-3=0

step2: factorise the equation.

(2n-1)(2n+3)=0

step3: make each bracket equal to zero.

2n-1=0

2n+3=0

step4: solve for n values.

1. 2n-1=0

(add 1 for both sides)

2n=1

(divide by 2 for both sides)

n= 1/2 ( $\frac{1}{2}$ ) or (0.5)

2. 2n+3=0

( subtract 3 for both sides)

2n=-3

(divide by 2 for both sides)

n= -3/2 ( $-\frac{3}{2}$ ) or (-1.5)

3 0

3 years ago

What is the reference angle for -144 degrees

Alexandra [31]

To find the reference angle for 144 you subtract 144 from 180 and that how you get your answer
180-144= 36*

6 0

4 years ago

A quality-control manager for a company that produces a certain soft drink wants to determine if a 12-ounce can of a certain bra

tensa zangetsu [6.8K]

Answer:

We conclude that the average calorie content of a 12-ounce can is greater than 120 calories.

Step-by-step explanation:

We are given that a quality-control manager for a company that produces a certain soft drink wants to determine if a 12-ounce can of a certain brand of soft drink contains 120 calories as the labeling indicates.

Using a random sample of 10 cans, the manager determined that the average calories per can is 124 with a standard deviation of 6 calories.

Let $\mu$ = average calorie content of a 12-ounce can.

So, Null Hypothesis, $H_0$ : $\mu \leq$ 120 calories {means that the average calorie content of a 12-ounce can is less than or equal to 120 calories}

Alternate Hypothesis, $H_A$ : $\mu$ > 120 calories {means that the average calorie content of a 12-ounce can is greater than 120 calories}

The test statistics that would be used here One-sample t test statistics as we don't know about the population standard deviation;

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

where, $\bar X$ = sample average calories per can = 124 calories

s = sample standard deviation = 6 calories

n = sample of cans = 10

So, test statistics = $\frac{124-120}{\frac{6}{\sqrt{10} } }$ ~ $t_9$

= 2.108

The value of t test statistics is 2.108.

Now, at 0.05 significance level the t table gives critical value of 1.833 at 9 degree of freedom for right-tailed test. Since our test statistics is more than the critical values of t as 2.108 > 1.833, so we have sufficient evidence to reject our null hypothesis as it will in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the average calorie content of a 12-ounce can is greater than 120 calories.

7 0

3 years ago

Segments AB , CD , and EF intersect at point O, points A, E, C and points B, F, D are collinear so that AO ≅ OB , CO ≅ OD . Prov

katovenus [111]

If points A, E and C are colinear, then they lie on the same line. The same statement you can say about points B, F and D.

1. Consider triangles AOC and BOD. In these triangles:

AO≅OB (given);
CO≅OD (given);
∠AOC≅∠BOD (as vertical angles).

Thus, ΔAOC≅ΔBOD by SAS Postulate (If any two corresponding sides and their included angle are the same in both triangles, then the triangles are congruent). Corresponding parts of congruent triangles are congruent, then

AC≅BD;
∠ACO≅∠BDO;
∠CAO≅∠DBO.

Since angles ACO and BDO are alternate interior angles between lines AE and BF with transversal CD and these angles are congruent, then lines AE and BF are parallel.

This gives you that

∠CEO≅∠OFD;
∠ECO≅∠ODF.

2. Consider triangles ECO and FDO. In these triangles

∠CEO≅∠OFD (previous proof);
CO≅OD (given);
∠ECO≅∠ODF (previous proof).

Therefore, ΔECO≅ΔFDO by AAS Postulate (if two angles and the non-included side one triangle are congruent to two angles and the non-included side of another triangle, then these two triangles are congruent). Then CE≅FD.

3. Note that

AE=AC+CE;
BF=BD+DF.

Since AC≅BD and CE≅DF, then AE=AC+CE=BD+DF=BF.

7 0

3 years ago