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3 years ago

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A freight train is half a mile long. It approaches a tunnel traveling 60mi/hr. If the tunnel is 1 mile long, how long will it ta

ke the train, at that speed, to pass completely through?

2 answers:

Marysya12 [62]3 years ago

8 0

Answer:

60min

Step-by-step explanation:

MrRa [10]3 years ago

6 0

Answer:60

Step-by-step explanation:I seen the answer on bill bye the math guy

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What is the value of x drawing not to scale

AVprozaik [17]

angle x would end up being equal to 71 degrees.

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3 years ago

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Can segments with lengths of 15, 20, and 36 form a triangle?

strojnjashka [21]

Ok, so remember

the legnth of the longest side must be LESS THAN the sum of the measures of the other 2 sides
if no longest side (becasue it has 2 longest sides or 3 equal sides), then it can form a triangle

so the longest side is 36

36 must be less than the sum of 15 and 20
36<15+20
36<35
false

therfor it cannot form a triangle
try it yourself, you cannot connect them that way

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3 years ago

Create a Sample Space diagram and work out the probability - There is a 4 sided dice (numbers 1 to 4) and an 8 sided dice (numbe

tester [92]

Answer:

7 / 16

Step-by-step explanation:

Sample space is attached below :

Theoretical Probability of any event A

P(A) = (number of required outcome / Total number of possible outcomes)

Required outcome = sum greater than 7 = 14

Total number of possible outcomes = 32

P(sum greater than 7) = 14 / 32

P(sum greater than 7) = 7 / 16

3 0

3 years ago

Guys how do you find out a average of some numbers?

blsea [12.9K]

Add all the number, and then divide it by number of numbers.
for example, 1, 2, 3, 7, 10
(1+2+3+7+10)/5
23/5 = 4.6 The average is 4.6

8 0

3 years ago

You are given the sample mean and the population standard deviation. Use this information to construct the 90% and 95% confide

salantis [7]

Answer:

With 90% confidence, it can be said that the population mean price lies in the first interval. With 95% confidence, it can be said that the population mean price lies in the second interval. The 95% confidence interval is wider than the 90%.

Step-by-step explanation:

We are given that a random sample of 60 home theater systems has a mean price of$131.00. Assume the population standard deviation is$18.80.

Firstly, the pivotal quantity for 90% confidence interval for the population mean is given by;

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

where, $\bar X$ = sample mean price = $131

$\sigma$ = population standard deviation = $18.80

n = sample of home theater = 60

$\mu$ = population mean

<em>Here for constructing 90% confidence interval we have used One-sample z test statistics as we know about the population standard deviation.</em>

<u>So, 90% confidence interval for the population mean, </u> $\mu$ <u> is ;</u>

P(-1.645 < N(0,1) < 1.645) = 0.90 {As the critical value of z at 5% level

of significance are -1.645 & 1.645}

P(-1.645 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.645) = 0.90

P( $-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.90

P( $\bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.90

<u>90% confidence interval for</u> $\mu$ = [ $\bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $131-1.645 \times {\frac{18.8}{\sqrt{60} } }$ , $131+1.645 \times {\frac{18.8}{\sqrt{60} } }$ ]

= [127.01 , 134.99]

Therefore, 90% confidence interval for the population mean is [127.01 , 134.99].

Now, the pivotal quantity for 95% confidence interval for the population mean is given by;

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

where, $\bar X$ = sample mean price = $131

$\sigma$ = population standard deviation = $18.80

n = sample of home theater = 60

$\mu$ = population mean

<em>Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about the population standard deviation.</em>

<u>So, 95% confidence interval for the population mean, </u> $\mu$ <u> is ;</u>

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

<u>95% confidence interval for</u> $\mu$ = [ $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $131-1.96 \times {\frac{18.8}{\sqrt{60} } }$ , $131+1.96 \times {\frac{18.8}{\sqrt{60} } }$ ]

= [126.24 , 135.76]

Therefore, 95% confidence interval for the population mean is [126.24 , 135.76].

Now, with 90% confidence, it can be said that the population mean price lies in the first interval. With 95% confidence, it can be said that the population mean price lies in the second interval. The 95% confidence interval is wider than the 90%.

7 0

3 years ago