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3 years ago

3/4+1/12 = simplest form

2 answers:

Studentka2010 [4]3 years ago

8 0

Your answer would be 5/6 in its simplest form.

To add fractions we must get them both over the same denominator, which means we can multiply 3/4 by 3 to get 9/12.

Then we can add 9/12 to 1/12 to get 10/12, which simplifies to 5/6 by dividing by 2.

I hope this helps!

jarptica [38.1K]3 years ago

4 0

Hey , here is the answer to ur question...!
3/4+1/2=5/4...
Hope this helps u...!

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The coordinates of the vertices of quadrilateral ABCD are A(−4, −1), B(−1, 2), C(5, 1), and D(1, −3).

serg [7]

Answer:

Sorry this is late, but it could help future people. Answers are in the picture.

Step-by-step explanation:

I took the quiz. :)

3 0

3 years ago

Read 2 more answers

Urgent answer needed pls

worty [1.4K]

Answer:

48 cm²

Step-by-step explanation:

PTRS is parallelogram with the base length= 8cm and the height as same as the height of ∆SUR.

the height = 2×24/8 = 6cm

so, the area of PTRS= 8×6 = 48 cm²

6 0

3 years ago

If a die is rolled one time, find probability of getting a number greater than 3 and an odd number.

Nadya [2.5K]

Answer:

$\dfrac{1}{6}$ .

Step-by-step explanation:

It is given that a die is rolled one time. So,

Sample space : $S=\{1,2,3,4,5,6\}$

Number is greater than 3 : $A=\{4,5,6\}$

Number is odd : $A=\{1,3,5\}$

Number greater than 3 and an odd number : $A\cap B=\{5\}$

It is conclude that,

$n(S)=6, n(A)=3,n(B)=3,n(A\cap B)=1$

The probability of getting a number greater than 3 and an odd number is

$P=\dfrac{n(A\cap B)}{n(S)}$

$P=\dfrac{1}{6}$

Therefore, the required probability is $\dfrac{1}{6}$ .

4 0

3 years ago

An article describes an experiment to determine the effectiveness of mushroom compost in removing petroleum contaminants from so

alexgriva [62]

Solution :

Let $p_1$ and $p_2$ represents the proportions of the seeds which germinate among the seeds planted in the soil containing $3\%$ and $5\%$ mushroom compost by weight respectively.

To test the null hypothesis $H_0: p_1=p_2$ against the alternate hypothesis $H_1:p_1 \neq p_2$ .

Let $\hat p_1, \hat p_2$ denotes the respective sample proportions and the $n_1, n_2$ represents the sample size respectively.

$$\hat p_1 = \frac{74}{155} = 0.477419$

$n_1=155$

$$p_2=\frac{86}{155}=0.554839$

$n_2=155$

The test statistic can be written as :

$$z=\frac{(\hat p_1 - \hat p_2)}{\sqrt{\frac{\hat p_1 \times (1-\hat p_1)}{n_1}} + \frac{\hat p_2 \times (1-\hat p_2)}{n_2}}}$

which under $H_0$ follows the standard normal distribution.

We reject $H_0$ at $0.05$ level of significance, if the P-value or if $|z_{obs}|>Z_{0.025}$

Now, the value of the test statistics = -1.368928

The critical value = $\pm 1.959964$

P-value = $P(|z|> z_{obs})= 2 \times P(z< -1.367928)$

$=2 \times 0.085667$

= 0.171335

Since the p-value > 0.05 and $|z_{obs}| \ngtr z_{critical} = 1.959964$ , so we fail to reject $H_0$ at $0.05$ level of significance.

Hence we conclude that the two population proportion are not significantly different.

Conclusion :

There is not sufficient evidence to conclude that the $\text{proportion}$ of the seeds that $\text{germinate differs}$ with the percent of the $\text{mushroom compost}$ in the soil.

8 0

3 years ago

A card is drawn from the deck. What is the probability it is either black face card or a<br> spade?

IRINA_888 [86]

Answer:

about 37% chance or, rounding to the nearest tenth, 36.5% chance

Step-by-step explanation:

since there are only 6 black face cards out of 52 cards, so the fraction is 6/52

there are also 13 spade cards out of 52 cards, so the fraction is 13/52

to find the percent you have to add the two fractions becuase all you want to do is draw a black face card or a spade card. when you add them you get 19/52

then dividing the fraction gives you .3653846154 on my calculator :)

4 0

4 years ago