PLEASE HELP MATH EXPAND LOG

A train station has installed a system for determining whether bags contain explosives. It has a 95% chance of correctly identif

astraxan [27]

Answer:

a) P(identified as containing explosives)=P(actually contains explosives and identified as containing explosives)+P(actually not contains explosives and identified as containing explosives)

=(10/(4*106))*0.95+(1-10/(4*106))*0.005 =0.005002363

hence probability that it actually contains explosives given identified as containing explosives)

=(10/(4*106))*0.95/0.005002363=0.000475

b)

let probability of correctly identifying a bag without explosives be a

hence a =0.99999763 ~ 99.999763%

c)

No as even if that becomes 1 ; proportion of true explosives will always be less than half of total explosives detected,

3 0

2 years ago

a sports statistician claim that the mean winning times for boston marathon women's open division champions is at least 2.68 hou

GenaCL600 [577]

Answer:

We conclude that the mean winning times for Boston marathon women's open division champions is at least 2.68 hours.

Step-by-step explanation:

We are given that a sports statistician claim that the mean winning times for Boston marathon women's open division champions is at least 2.68 hours.

The mean winning time of a sample of 30 randomly selected Boston marathon women's open division champions is 2.60 hours. assume the population standard deviation is 0.32 hours.

Let $\mu$ = mean winning times for Boston marathon women's open division champions.

So, Null Hypothesis, $H_0$ : $\mu\geq$ 2.68 hours {means that the mean winning times for Boston marathon women's open division champions is at least 2.68 hours}

Alternate Hypothesis, $H_A$ : p < 2.68 hours {means that the mean winning times for Boston marathon women's open division champions is less than 2.68 hours}

The test statistics that would be used here One-sample z test statistics as we know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean winning time = 2.60 hours

$\sigma$ = population standard deviation = 0.32 hours

n = sample of women's open division champions = 30

So, test statistics = $\frac{2.60-2.68}{\frac{0.32}{\sqrt{30} } }$

= -1.37

The value of z test statistics is -1.37.

Now, P-value of the test statistics is given by following formula;

P-value = P(Z < -1.37) = 1 - P(Z $\leq$ 1.37)

= 1 - 0.9147 = 0.0853

Since, P-value of the test statistics is more than the level of significance as 0.0853 > 0.05, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the mean winning times for Boston marathon women's open division champions is at least 2.68 hours.

6 0

2 years ago

The rectangular playground in tim's school is three times as long as it is wide. the area of the playground is 75 square meters.

qwelly [4]

Let x be the width of the playground, then 3x is the length of the
playground

х * 3х = 75
3x² = 75
x² = 25
x = 5 m (width)
5*3=15 m (length)

Perimeter = 2(5+15) = 2*20 = 40 meters.

8 0

2 years ago

Consider the graph of the linear function h(x) = –x + 5. Which could you change to move the graph down 3 units?

KATRIN_1 [288]

The +5 because that's the y value on the graph

8 0

2 years ago

Read 2 more answers

a video game arcade offers a yearly membership with reduced rates for game play. A single membership costs $60 per year. Game to

7nadin3 [17]

slope is $0.10 ($1.00 per 10 tokens = $0.10)

y-intercept is $60 ($60 is the annual fee)