PLEASE HELP ASAP!!!!!!!!!!!!!!!!!!!!!!

While standing on a highway overpass, Jennifer wonders what proportion of the vehicles that pass on the highway below are trucks

Assoli18 [71]

Answer:Obtain a systematic sample by selecting every 20th vehicle that passes (in any lane and going in any direction).

Step-by-step explanation:

8 0

3 years ago

What is the value of y?

Sati [7]

I need more information for this, the y-axis!

4 0

3 years ago

Read 2 more answers

The lengths of two sides of a triangle are 5 and

Gwar [14]

Answer:

D) 2

Step-by-step explanation:

A Triangle two sides added together must always be greater than the other side.

It can not be 2 because the two sides of a triangle added are not greater than the other side.

Here 5 + 2 = 7 which is not greater than 7 on the other side, so 2 fails.

5 0

2 years ago

Sixty percent means the same as _______.

ycow [4]

60% - 3/5
Hope this help, don't really know what your expecting the answer to be

3 0

3 years ago

A laboratory scale is known to have a standard deviation (sigma) or 0.001 g in repeated weighings. Scale readings in repeated we

weqwewe [10]

Answer:

99% confidence interval for the given specimen is [3.4125 , 3.4155].

Step-by-step explanation:

We are given that a laboratory scale is known to have a standard deviation (sigma) or 0.001 g in repeated weighing. Scale readings in repeated weighing are Normally distributed with mean equal to the true weight of the specimen.

Three weighing of a specimen on this scale give 3.412, 3.416, and 3.414 g.

Firstly, the pivotal quantity for 99% confidence interval for the true mean specimen is given by;

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

where, $\bar X$ = sample mean weighing of specimen = $\frac{3.412+3.416+3.414}{3}$ = 3.414 g

$\sigma$ = population standard deviation = 0.001 g

n = sample of specimen = 3

$\mu$ = population mean

<em>Here for constructing 99% confidence interval we have used z statistics because we know about population standard deviation (sigma).</em>

So, 99% confidence interval for the population mean, $\mu$ is ;

P(-2.5758 < N(0,1) < 2.5758) = 0.99 {As the critical value of z at 0.5% level

of significance are -2.5758 & 2.5758}

P(-2.5758 < $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ < 2.5758) = 0.99

P( $-2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X - \mu}$ < $2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.99

P( $\bar X-2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.99

<u>99% confidence interval for</u> $\mu$ = [ $\bar X-2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $3.414-2.5758 \times {\frac{0.001}{\sqrt{3} } }$ , $3.414+2.5758 \times {\frac{0.001}{\sqrt{3} } }$ ]

= [3.4125 , 3.4155]

Therefore, 99% confidence interval for this specimen is [3.4125 , 3.4155].

6 0

3 years ago