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4 years ago

What is the distance between –3 and 2 on the number line?

Mathematics

2 answers:

Dennis_Churaev [7]4 years ago

5 0

Answer: 5

Step-by-step explanation: The distance between two numbers can be found by finding the absolute value of the difference of the numbers. Put more simply, subtract -3 - 2 to get -5, which has an absolute value of 5. Or, you can do 2 - -3 to get 5, which has the same absolute value of 5.

Neporo4naja [7]4 years ago

5 0

Answer:

Step-by-step explanation:

first you add the negative into the postive , its turns to a possitive 3 than you ad it with 2 then get( 5)

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Aleonysh [2.5K]

Answer:

He would run a half of a lap because if he ran a full lap in 2 minutes than in 1 minute it would be half of 1.

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

23% of $640 Please show work

Tema [17]

23/100×$640=147.2 is the answer

3 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

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baherus [9]

Answer:

Step-by-step explanation:

4 0

3 years ago

There is a 10% chance of the mean oil-change time being at or below the sample mean for which there is an area of 0.10 to the

Zarrin [17]

The probability that a normally distributed data with a mean of μ and a standard deviation of σ is less than a value x is given by:

$P(X\ \textless \ x)=P\left(z\ \textless \ \frac{x-\mu}{\sigma} \right)$

Given that μ = 16.5 and σ = 0.804984 and that the probability that the mean oil-change time being at or below the sample mean for which there is an area of 0.10 to the left under the normal curve, then:

$P\left(z\ \textless \ \frac{x-16.5}{0.804984} \right)=0.1\\ \\ \Rightarrow P\left(z\ \textless \ \frac{x-16.5}{0.804984} \right)=P(z\ \textless \ -1.281) \\ \\ \Rightarrow\frac{x-16.5}{0.804984}=-1.281 \\ \\ \Rightarrow x-16.5=-1.281(0.804984)=-1.031 \\ \\ x=-1.031+16.5=15.47$

6 0

4 years ago