What is the absolute value of 24

Cable Strength: A group of engineers developed a new design for a steel cable. They need to estimate the amount of weight the ca

KatRina [158]

Answer:

95% confidence interval for the mean breaking strength of the new steel cable is [763.65 lb , 772.75 lb].

Step-by-step explanation:

We are given that the engineers take a random sample of 45 cables and apply weights to each of them until they break. The mean breaking weight for the 45 cables is 768.2 lb. The standard deviation of the breaking weight for the sample is 15.1 lb.

Since, in the question it is not specified that how much confidence interval has be constructed; so we assume to be constructing of 95% confidence interval.

Firstly, the Pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean breaking weight = 768.2 lb

s = sample standard deviation = 15.1 lb

n = sample of cables = 45

$\mu$ = population mean breaking strength

Here for constructing 95% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-2.02 < $t_4_4$ < 2.02) = 0.95 {As the critical value of t at 44 degree

of freedom are -2.02 & 2.02 with P = 2.5%}

P(-2.02 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.02) = 0.95

P( $-2.02 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.02 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-2.02 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.02 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-2.02 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.02 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $768.2-2.02 \times {\frac{15.1}{\sqrt{45} } }$ , $768.2+2.02 \times {\frac{15.1}{\sqrt{45} } }$ ]

= [763.65 lb , 772.75 lb]

Therefore, 95% confidence interval for the mean breaking strength of the new steel cable is [763.65 lb , 772.75 lb].

3 0

4 years ago

Consider the functions shown below.

I am Lyosha [343]

Answer: D

Step-by-step explanation:

A. False

f(x): x → -∞, y → -∞

g(x): x → -∞, y → 1

B. False

at x = -1, f(x) = -7

at x = -1, g(x) = 1.5

C. False

refer to A

g(x) → 1 and f(x) → -∞ so g(x) > f(x)

D. TRUE

It is hard to tell from the graph so plug in a value to check.

f(10) = 5(10) - 2

= 50 - 2

= 48

g(10) = 2¹⁰ + 1

= 1024 + 1

= 1025

g(10) > f(10)

6 0

3 years ago

Combine like radicals to get your answer. image attached

Lena [83]

I believe it’s -6 square root of 2 + 3 square root of 4 + 6 square root of 5

3 0

4 years ago

When is it most handy to use a table to display information? When might you use a graph instead?

Sunny_sXe [5.5K]

Answer:

Use a table

The display will be used to look up individual values.
It will be used to compare individual values but not entire series of values to one another.
Precise values are required.
The quantitative information to be communicated involves more than one unit of measure.
Both summary and detail values are included.

Use a graph

The message is contained in the shape of the values (e.g. patterns, trends, exceptions).
The display will be used to reveal relationships among whole sets of values.

Source

In the comments

Hope this helps!

3 0

3 years ago

How to solve question 19 thanks

miskamm [114]

Let $X$ be the random variable for the weight of any given can, and let $\mu$ and $\sigma$ be the mean and standard deviation, respectively, for the distribution of $X$ .

You have

$\begin{cases}\mathbb P(X377)=0.025\end{cases}$

Recall that for any normal distribution, approximately 99.7% of it lies within three standard deviations of the mean, i.e. $\mathbb P(\mu-3\sigma$ . This means 0.3% must lie outside this range, $\mathbb P(X\mu+3\sigma)=0.003$ . Because the distribution is symmetric, it follows that $\mathbb P(X$ .

Also recall that for any normal distribution, about 95% of it falls within two standard deviations of the mean, so $\mathbb P(\mu-2\sigma$ , which means 5% falls outside, and by symmetry, $\mathbb P(X>\mu+2\sigma)=0.025$ .

Together this means

$\begin{cases}\mu-3\sigma=347\\\mu+2\sigma=377\end{cases}$

Solving for the mean and standard deviation gives $\mu=365$ and $\sigma=6$ .

6 0

4 years ago