If c = 4 and d = 5, find c:d. 1:5 4:5 5:4

The quality control manager of a chemical company randomly sampled twenty 100-pound bags of fertilizer to estimate the variance

lisabon 2012 [21]

Answer:

The 95% confidence interval for the variance in the pounds of impurities would be $3.829 \leq \sigma^2 \leq 14.121$ .

Step-by-step explanation:

1) Data given and notation

$s^2 =6.62$ represent the sample variance

s=2.573 represent the sample standard deviation

$\bar x$ represent the sample mean

n=20 the sample size

Confidence=95% or 0.95

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population mean or variance lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

The Chi square distribution is the distribution of the sum of squared standard normal deviates .

2) Calculating the confidence interval

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

On this case the sample variance is given and for the sample deviation is just the square root of the sample variance.

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

$df=n-1=20-1=19$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical values.

The excel commands would be: "=CHISQ.INV(0.025,19)" "=CHISQ.INV(0.975,19)". so for this case the critical values are:

$\chi^2_{\alpha/2}=32.852$

$\chi^2_{1- \alpha/2}=8.907$

And replacing into the formula for the interval we got:

$\frac{(19)(6.62)}{32.852} \leq \sigma^2 \frac{(19)(6.62)}{8.907}$

$3.829 \leq \sigma^2 \leq 14.121$

So the 95% confidence interval for the variance in the pounds of impurities would be $3.829 \leq \sigma^2 \leq 14.121$ .

4 0

3 years ago

Express as a sum

zmey [24]

Answer:

I think it would be

Sums

1) a + -5

2) m + -n

Differences

1) 7 - -3

2) a - -5

Step-by-step explanation:

6 0

2 years ago

A ball is thrown into the air with an upward velocity of 36 ft/s. Its height h in feet after t seconds is given by the function

elixir [45]

This can be solved in two ways: With heavy tools or with just algebra.
What is your level? Have you studied calculus?

With pure algebra:
We need to find the maximum of the function h = −16t^2 + 36t + 5
Lets take out -1 for simplicity:
h = −(16t2 - 36t - 5)
For now lets just work with this: 16t^2 - 36t - 5
16t^2=(4t)^2
(4t-x)^2= 16t^2-2*4xt+x^2
we have -36t so x should be 4.5 as 2*4*4.5=36

Lets see what we have now:
16t^2 - 36t - 5= (4t-4.5)^2 is this true? No but close
(4t-4.5)^2= 16t^2- 2*4*4.5t +4.5^2= 16t^2-36t+20.25

16t^2 - 36t - 5 and 16t^2-36t+20.25 nearl the same just take away 25.25 from the right hand side

Getting long, just stay with me:
16t^2 - 36t - 5= (4t-4.5)^2 - 25.25

h= -{(4t-4.5)^2 -25.25}
h=-(4t-4.5)^2 + 25.25

We want to find the maximum of this function. -(4t-4.5)^2 this bit is always negative or 0, so it maximum is when it is 0. Solve: 4t-4.5=0
t=1,125

7 0

3 years ago

The legs of a right triangle are 18 centimeters and 80 centimeters long. What is the length of the hypotenuse?

noname [10]

A^2 + b^2 = c^2....a and b are ur legs and c is ur hypotenuse
18^2 + 80^2 = c^2
324 + 6400 = c^2
6724 = c^2
sq rt 6724 = c
82 = c.....so the hypotenuse is 82 cm

7 0

3 years ago

Read 2 more answers

Chocolate chip cookies have a distribution that is approximately normal with a mean of 23.6 chocolate chips per cookie and a sta

nikklg [1K]

Answer:

Those values can be helpul to find an ideal range for the number of chocolate chips per cookie.

P(5) = 19.82 chips per cookie.

P(95) = 27.38 chips per cookie.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 23.6, \sigma = 2.3$