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3 years ago

I don't get this... Please help.

Mathematics

2 answers:

VLD [36.1K]3 years ago

8 0

In this problem, you are just being asked the distance between the numbers on the number line.

Example:

X=5, A= - 5

Count the spaces in between the 2 numbers... 5 is on the positive side, any number on the right of a number line is a positive number, and any number on the left of a number line is a negative.
so if you count down from positive 5 to 0, you have jest then moved 5 spaces backwards... now move back 5 again into the negative side, now you should have moved a total of 10 spaces coming from positive to negative.

marin [14]3 years ago

5 0

The first one is c
the second one is b
the third one is a
the fourth one is b

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A square has a side length of 5ft what is its area

Citrus2011 [14]

Answer:

25 feet squared

Step-by-step explanation:

Formula for the area of a square:

A=s^2

5^2, or 5*5, equals 25.

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3 years ago

Which expression would you use to estimate 36% of 84? A. 30% of 84 B. 25% of 84 C. 40% of 84 D. 35% of 84 ANSWER IT FOR ME PLEAS

ludmilkaskok [199]

Answer:

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Step-by-step explanation

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3 years ago

Simplify the expression 4*4+4*x

leva [86]

Answer: The answer is 16+4x

Step-by-step explanation:

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3 years ago

luis has three more quarters than dimes in his pocket, for a total of $1.80. How many dimes (x) are in his pocket

Dmitry_Shevchenko [17]

Answer:

Step-by-step explanation:

In this case, Luis has 6 quarters, and 3 dimes. 6 - 3 = 3, and it works out to the following:

6 • 0.25 = 1.50

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8 0

3 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