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3 years ago

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A test subject is randomly selected for a pregnancy test. What is the probability of getting a subject who is not pregnant, give

n that the test result is positive? Find the probability using the data table.
Positive Test Negative Test
Subject is pregnant. 78 7
Subject is not pregnant. 12 28

A. 0.10

B. 0.13

C. 0.19

D. 0.25

2 answers:

vitfil [10]3 years ago

4 0

You're given that the test result is positive, so only look at the first column not pregnant = 12 total = 78+12 = 90 probability = 12/90 = 0.13333333.

Hence, the ans would be B.

Minchanka [31]3 years ago

3 0

The correct answer is
B) 0.13

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Answer:

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Step-by-step explanation:

$-10+\left|6+\left(-17\right)\right|-\left(-14\right)$
$=-10+\left|6-17\right|+14$
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2 years ago

(2.5×10^−10)(7×10^−6)

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<span>(2.5×10^−10)(7×10^−6)
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4 years ago

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Answer:

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3 years ago

Read 2 more answers

The length of a rectangle is 5 metres less than twice the breadth. If the perimeter is 50 meters,find the length and breadth

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<h3><u>S</u><u> </u><u>O</u><u> </u><u>L</u><u> </u><u>U</u><u> </u><u>T</u><u> </u><u>I</u><u> </u><u>O</u><u> </u><u>N</u><u> </u><u>:</u></h3>

As per the given question, it is stated that the length of a rectangle is 5 m less than twice the breadth.

Assumption : Let us assume the length as "l" and width as "b". So,

$\twoheadrightarrow \quad\sf{ Length =2(Width)-5}$

$\twoheadrightarrow \quad\sf{ \ell=(2b-5) \; m}$

Also, we are given that the perimeter of the rectangle is 50 m. Basically, we need to apply here the formula of perimeter of rectangle which will act as a linear equation here.

$\\ \twoheadrightarrow \quad\sf{ Perimeter_{(Rectangle)} = 2(\ell +b) } \\$

<em>l</em> denotes length
<em>b</em> denotes breadth

$\\ \twoheadrightarrow \quad\sf{50= 2(2b-5+b)} \\$

$\\ \twoheadrightarrow \quad\sf{50= 2(3b-5)} \\$

$\\ \twoheadrightarrow \quad\sf{50= 6b - 10} \\$

$\\ \twoheadrightarrow \quad\sf{50+10= 6b} \\$

$\\ \twoheadrightarrow \quad\sf{60= 6b} \\$

$\\ \twoheadrightarrow \quad\sf{\cancel{\dfrac{60}{6}}=b} \\$

$\\ \twoheadrightarrow \quad\underline{\bf{10\; m = Width }} \\$

Now, finding the length. According to the question,

$\twoheadrightarrow \quad\sf{ \ell=(2b-5) \; m}$

$\twoheadrightarrow \quad\sf{ \ell=2(10)-5\; m}$

$\twoheadrightarrow \quad\sf{ \ell=20-5\; m}$

$\\ \twoheadrightarrow \quad\underline{\bf{15\; m = Length }} \\$

<u>Therefore</u><u>,</u><u> </u><u>length</u><u> </u><u>and</u><u> </u><u>breadth</u><u> </u><u>of</u><u> </u><u>the</u><u> </u><u>r</u><u>ectangle</u><u> </u><u>is</u><u> </u><u>1</u><u>5</u><u> </u><u>m</u><u> </u><u>and</u><u> </u><u>10</u><u> </u><u>m</u><u>.</u><u> </u>

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3 years ago

2. The National Safety Council routinely analyzes the benefit of seat belt use on driver safety. Their data showed that among 28

JulijaS [17]

Answer:

We conclude that there is difference in the proportion of deaths between the 2 groups.

Step-by-step explanation:

We are given that among 2823 drivers not wearing seat belts, 31 died as a result of injuries, and among 7765 drivers wearing seat belts 16 were killed.

Let $p_1$ = <u><em>proportion of deaths when drivers were not wearing seat belts.</em></u>

$p_2$ = <u><em>proportion of deaths when drivers were wearing seat belts.</em></u>

So, Null Hypothesis, $H_0$ : $p_1=p_2$ {means that there is no difference in the proportion of deaths between the 2 groups}

Alternate Hypothesis, $H_A$ : $p_1\neq p_2$ {means that there is difference in the proportion of deaths between the 2 groups}

The test statistics that would be used here <u>Two-sample z test for proportions;</u>

T.S. = $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ ~ N(0,1)

where, $\hat p_1$ = sample proportion of deaths when drivers were not wearing seat belts = $\frac{31}{2823}$ = 0.011

$\hat p_2$ = sample proportion of deaths when drivers were wearing seat belts = $\frac{16}{7765}$ = 0.002

$n_1$ = sample of drivers not wearing seat belts = 2823

$n_2$ = sample of drivers wearing seat belts = 7765

So, <u><em>the test statistics</em></u> = $\frac{(0.011-0.002)-(0)}{\sqrt{\frac{0.011(1-0.011)}{2823}+\frac{0.002(1-0.002)}{7765} } }$

= 4.438

The value of z test statistics is 4.438.

<u>Now, at 5% significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.</u>

Since our test statistic doesn't lie within the range of critical values of z, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <u>we reject our null hypothesis</u>.

Therefore, we conclude that there is difference in the proportion of deaths between the 2 groups.

Also, <u>Margin of error</u> (E) = $1.96 \times \sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} }$

= $1.96 \times \sqrt{\frac{0.011(1-0.011)}{2823}+\frac{0.002(1-0.002)}{7765} }$

= <u>0.00397</u>

5 0

3 years ago