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2 years ago

14

Can somebody please help me with the question below and please explain how to find the answer? plz, I'm studying for a test plzz

zz help!!!

1 answer:

blagie [28]2 years ago

3 0

Answer:

-10/9 < -2/3 < 4/6 < 6/5 < 5/4

(hope this helps!)

Step-by-step explanation:

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When a truckload of apples arrives at a packing plant, a random sample of 150 is selected and examined for bruises, discoloratio

myrzilka [38]

Answer:

c. 0.088

Step-by-step explanation:

Let p(s) be the proportion of apples defected in the sample. The probability that p(s)<0.05 can be calculated by calculating z statistic of 0.05:

$\frac{p(s)-p}{\sqrt{\frac{p*(1-p)}{N} } }$ where

p(s) = 0.05

p is the proportion of the apples in fact defected (0.08)

N is the sample size (150) Then,

z(0.05)= $\frac{0.05-0.08}{\sqrt{\frac{0.08*0.92)}{150} } }=-1.354$

And P(z<-1.354)≈0.0879

Therefore the probability that a random sample of 150 among 8% defected apples can be accepted is 0.088.

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3 years ago

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statuscvo [17]

Holy guacamole....oml

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2 years ago

A tip in a restaurant has been typically figured at 15% of the total bill.

Anastasy [175]

Answer:

4.5 dollars (before tax id assume)

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3 years ago

Giving brainliest!???!!!!!

Jlenok [28]

Answer: B is the answer

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3 years ago

Read 2 more answers

In a completely randomized experimental design, three brands of paper towels were tested for their ability to absorb water. Equa

arlik [135]

Answer:

Yes. At this significance level, there is evidence to support the claim that there is a difference in the ability of the brands to absorb water.

Step-by-step explanation:

<em>The question is incomplete:</em>

<em>The significance level is 0.05.</em>

<em>The data is:</em>

<em>Brand X: 91, 100, 88, 89</em>

<em>Brand Y: 99, 96, 94, 99</em>

<em>Brand Z: 83, 88, 89, 76</em>

We have to check if there is a significant difference between the absorbency rating of each brand.

Null hypothesis: all means are equal

$H_0:\mu_x=\mu_y=\mu_z$

Alternative hypothesis: the means are not equal

$H_a: \mu_x\neq\mu_y\neq\mu_z$

We have to apply a one-way ANOVA

We start by calculating the standard deviation for each brand:

$s_x^2=30,\,\,s_y^2=6,\,\,s_z^2=35.33$

Then, we calculate the mean standard error (MSE):

$MSE=(\sum s_i^2)/a=(30+6+35.33)/3=71.33/3=23.78$

Now, we calculate the mean square between (MSB), but we previously have to know the sample means and the mean of the sample means:

$M_x=92,\,\,M_y=97,\,\,M_z=84\\\\M=(92+97+84)/3=91$

The MSB is then:

$s^2=\dfrac{\sum(M_i-M)^2}{N-1}\\\\\\s^2=\dfrac{(92-91)^2+(97-91)^2+(84-91)^2}{3-1}\\\\\\s^2=\dfrac{1+36+49}{2}=\dfrac{86}{2}=43\\\\\\\\MSB=ns^2=4*43=172$

Now we calculate the F statistic as:

$F=MSB/MSE=172/23.78=7.23$

The degrees of freedom of the numerator are:

$dfn=a-1=3-1=2$

The degrees of freedom of the denominator are:

$dfd=N-a=3*4-3=12-3=9$

The P-value of F=7.23, dfn=2 and dfd=9 is:

$P-value=P(F>7.23)=0.01342$

As the P-value (0.013) is smaller than the significance level (0.05), the null hypothesis is rejected.

There is evidence to support the claim that there is a difference in the ability of the brands to absorb water.

3 0

3 years ago