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3 years ago

Show work and Evaluate 3/4 + 2/5

Mathematics

1 answer:

frez [133]3 years ago

5 0

Answer:

23/20

Step-by-step explanation:

Find the common denominator of the 2 fractions (It's 20):

3/4 + 2/5

Multiply both numerator and denominator by 5 for the first fraction and multiply by 4 for the second fraction:

15/20 + 8/20

Add the fractions together:

23/20

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I need help help plz help

Leni [432]

Answer:

16 (The last one)

Step-by-step explanation:

All of them equal 1/16 OTHER than the last one.

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3 years ago

4px = 20 (find p) -4px = 20 (find x)

Kazeer [188]

Answer:

4px=20

x=(20/4p)-----(1)

-4px=20--------(2)

putting value of x in epn 2

-4p*20=20*4p

-80p=+80p

Both cut these value are rejected

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3 years ago

Prediction: It rains 2 out of the last 12 days in March. If this trend continues, how many rainy days would you expect in April?

Vladimir79 [104]

Answer:

I would expect rain 5 days

Step-by-step explanation:

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8 0

3 years ago

Alexxx [7]

Answer:

(a) P = 4x+10

(b) x = 6

Step-by-step explanation:

(a) The perimeter is the sum of the side lengths:

P = AB +AC +BC = 2x +2x +10

P = 4x+10

(b) For P=34, the value of x is ...

34 = 4x +10 . . . .substitute given value for P

24 = 4x . . . . . . . subtract 10

6 = x . . . . . . . . . .divide by 4

7 0

4 years ago

A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 400 babies were born, a

Masja [62]

Answer:

(a) 99% confidence interval for the percentage of girls born is [0.804 , 0.896].

(b) Yes, the proportion of girls is significantly different from 0.50.

Step-by-step explanation:

We are given that a clinical trial tests a method designed to increase the probability of conceiving a girl.

In the study 400 babies were born, and 340 of them were girls.

(a) Firstly, the pivotal quantity for 99% confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of girls born = $\frac{340}{400}$ = 0.85

n = sample of babies = 400

p = population percentage of girls born

Here for constructing 99% confidence interval we have used One-sample z proportion statistics.

So, 99% confidence interval for the population proportion, p is ;

P(-2.58 < N(0,1) < 2.58) = 0.99 {As the critical value of z at 0.5% level

of significance are -2.58 & 2.58}

P(-2.58 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 2.58) = 0.99

P( $-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

P( $\hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

99% confidence interval for p = [ $\hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.85-2.58 \times {\sqrt{\frac{0.85(1-0.85)}{400} } }$ , $0.85+2.58 \times {\sqrt{\frac{0.85(1-0.85)}{400} } }$ ]

= [0.804 , 0.896]

Therefore, 99% confidence interval for the percentage of girls born is [0.804 , 0.896].

(b) Let p = population proportion of girls born.

So, Null Hypothesis, $H_0$ : p = 0.50 {means that the proportion of girls is equal to 0.50}

Alternate Hypothesis, $H_A$ : p $\neq$ 0.50 {means that the proportion of girls is significantly different from 0.50}

The test statistics that will be used here is One-sample z proportion test statistics;

T.S. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of girls born = $\frac{340}{400}$ = 0.85

n = sample of babies = 400

So, the test statistics = $\frac{0.85-0.50}{\sqrt{\frac{0.85(1-0.85)}{400} } }$

= 19.604

Now, at 0.01 significance level, the z table gives critical value of 2.3263 for right tailed test. Since our test statistics is way more than the critical value of z as 19.604 > 2.3263, so we have sufficient evidence to reject our null hypothesis due to which we reject our null hypothesis.

Therefore, we conclude that the proportion of girls is significantly different from 0.50.

8 0

3 years ago