Combining Like Terms (CLT) can only happen when the variable parts are __

Shoe bargain corner types colors walking black basketball blue cross-trainer red badminton white baseball silver you notice the

Murljashka [212]

Baseball shoes are 1/5 of the shoe types. Red, white, or silver are 3/5 of the shoe colors. The probability of choosing a red, white, or silver baseball shoe is
(1/5)(3/5) = 3/25 = 0.12

3 0

3 years ago

A random sample of 20 students yielded a mean of ¯x = 72 and a variance of s2 = 16 for scores on a college placement test in mat

Helen [10]

Answer:

The 98% confidence interval for the variance in the pounds of impurities would be $8.400 \leq \sigma^2 \leq 39.827$ .

Step-by-step explanation:

1) Data given and notation

$s^2 =16$ represent the sample variance

s=4 represent the sample standard deviation

$\bar x$ represent the sample mean

n=20 the sample size

Confidence=98% or 0.98

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:

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

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 0.98 or 98%, the value of $\alpha=0.02$ and $\alpha/2 =0.01$ , and we can use excel, a calculator or a table to find the critical values.

The excel commands would be: "=CHISQ.INV(0.01,19)" "=CHISQ.INV(0.99,19)". so for this case the critical values are:

$\chi^2_{\alpha/2}=36.191$

$\chi^2_{1- \alpha/2}=7.633$

And replacing into the formula for the interval we got:

$\frac{(19)(16)}{36.191} \leq \sigma^2 \leq \frac{(19)(16)}{7.633}$

$8.400 \leq \sigma^2 \leq 39.827$

So the 98% confidence interval for the variance in the pounds of impurities would be $8.400 \leq \sigma^2 \leq 39.827$ .

7 0

4 years ago

Bri and Annamarie are picking strawberries at two different strawberry farms that charge a flat fee plus a cost per basket picke

denpristay [2]

The correct answer is a & d.
If you find the equation for Annamarie, you get y = 2.75x+ 11.

Using the slope formula to find what Annamarie pays per basket.
(38.5 - 11)/(10 - 0) = 2.75
Annamarie's flat fee is $11 and Bri's flat fee is $5.50

6 0

3 years ago

Read 2 more answers

Suppose Japan works for 6 days......

Nesterboy [21]

Answer:

B im sorry if I get it wrong.

Step-by-step explanation:

6 0

3 years ago

Philips Semiconductors is a leading European manufacturer of integrated circuits. Integrated circuits are produced on silicon wa

shepuryov [24]

Answer:

Step-by-step explanation:

1) The null hypothesis is,

H_0: The mean thickness of teh wafers for the five positions are equal

i.e, $H_0:\mu_1=\mu_2=\mu_3=\mu_4=\mu_5$

2)

The alternative hypothesis is,

H_1: There is an evidence of a difference in the mean thickness of the wafers for the five positions

3)

Let us consider the level of significance $\alpha=0.01$

from the Minitab outout

One-way ANOVA:C1 versus C2

source DF SS MS F P

C2 4 1417.73 354.43 51.00 0.00

Error 145 1007.77 6.95

Total 14 2425.50

S = 2.636 R - S = 58.45% R - Sq(adj) = 57.31%

Individual 95% CIs For Mean Based on Pooled StDev

level N Mean StDev -,----------,----------,----------,----------

1 30 240.53 2.62 (--,--)

2 30 243.73 2.79 (--,--)

3 30 246.07 2.90 (--,--)

4 30 249.10 2.66 (--,--)

5 30 247.07 2.15 (--,--)

-,----------,----------,----------,----------

240.0 243.0 246.0 249.0

Pooled StDev = 2.64

4)

The test statistic is, F = 51

5)

The P-value is approximately 0

6)

Here, the P - value is less than the level of significance

$\therefore \,P-value$

So, we do not accept our null hypothesis H_0

7)

Therefore, we conclude that there is an evidence of a difference in the mean thickness of the wafers for the five positions at level of significance $\alpha=0.05$

b)

chek attachment

we observe that,

The mean thickness of the wafer for position 1 is significant with position 2,

position 18, position 19 and position 28.

The mean thickness of the wafer for position 2 is significant with position 18,

position 19 and position 28.

The mean thickness of the wafer for position 18 is significant with the

position 19.

But the mean thickness of the wafer for position18 is not significant with

position 28.

The mean thickness of the wafer for position 19 is significant with position 28.

7 0

3 years ago

Combining Like Terms (CLT) can only happen when the variable parts are ___?