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3 years ago

The sum of three consecutive even integers is –78

Mathematics

1 answer:

Natasha_Volkova [10]3 years ago

7 0

Answer:

24, 26, 28

Step-by-step explanation:

So the three consecutive even numbers become 'x', 'x+2′, 'x+4′. According to the given condition, the sum of three consecutive even numbers is 78. x=24. Hence the three consecutive even numbers are 24, 26 and 28. is this the answer you were looking for?

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A scientist used 786 milliliters of a liquid for an experiment. How many liters of the liquid did the scientist use for this exp

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1/2 of $20.00 is $10.00

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Answer:

yes

Step-by-step explanation:

bc 20/2 is another way to fimd it which gets you 10

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Help please!!!!!! 12-13 8th grade math!!!

marin [14]

Answer:

12.

m∠ACB = 180° - 58° - 78° = 44°

m∠DCE = 180° - 85° - 60° = 35°

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13.

Base on the picture, we know that:

x° + 2x° + 3x° = 180°

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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.

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3 years ago

10ac x 6ab x (-2abc) = ?

blsea [12.9K]

Answer:

= -120a³b²c²

Step-by-step explanation:

10ac * 6ab * -2abc

10*6*-2 = -120

ac * ab * abc = a*a*a *b*b*c*c = a³b²c²

then:

10ac * 6ab * -2abc = -120* a³b²c²

= -120a³b²c²

3 0

4 years ago