I need help with this problem please

A sample of 4 different calculators is randomly selected from a group containing 18 that are defective and 35 that have no defec

Sphinxa [80]

<span>1.

P(</span>at least one of the calculators is defective)=

1- P(none of the selected calculators is defective).

2.

P(none of the selected calculators is defective)

=n(ways of selecting 4 non-defective calculators)/n(total selections of 4)

3.

selecting 4 non-defective calculators can be done in C(35, 4) many ways,

where $C(35, 4)= \frac{35!}{4!31!}= \frac{35*34*33*32*31!}{4!*31!}= \frac{35*34*33*32}{4!}= \frac{35*34*33*32}{4*3*2*1}= 52,360$

while, the total number selections of 4 out of 18+35=53 calculators can be done in C(53, 4) many ways,

$C(53, 4)= \frac{53!}{4!*49!}= \frac{53*52*51*50}{4*3*2*1}= 292,825$

4. so, P(none of the selected calculators is defective)= $\frac{52,360}{292,825} =0.18$

5. P(at least one of the calculators is defective)=

1- P(none of the selected calculators is defective)=1-0.18=0.82

Answer:0.82

7 0

3 years ago

A student at a four-year college claims that mean enrollment at four-year colleges is higher than at two-year colleges in the Un

WINSTONCH [101]

Answer:

Part 1: The statistic

$t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}$ (1)

And the degrees of freedom are given by $df=n_1 +n_2 -2=35+35-2=68$

Replacing we got

$t=\frac{(5135-4436)-0}{\sqrt{\frac{783^2}{35}+\frac{553^2}{35}}}}=4.31$

Part 2: P value

Since is a right tailed test the p value would be:

$p_v =P(t_{68}>4.31)=0.000022 \approx 0.00002$

Comparing the p value we see that is lower compared to the significance level of 0.01 so then we can reject the null hypothesis and we can conclude that the mean for the four year college is significantly higher than the mean for the two year college and then the claim makes sense

Step-by-step explanation:

Data given

$\bar X_{1}=5135$ represent the mean for four year college

$\bar X_{2}=4436$ represent the mean for two year college

$s_{1}=783$ represent the sample standard deviation for four year college

$s_{2}=553$ represent the sample standard deviation two year college

$n_{1}=35$ sample size for the group four year college

$n_{2}=35$ sample size for the group two year college

$\alpha=0.01$ Significance level provided

t would represent the statistic (variable of interest)

System of hypothesis

We need to conduct a hypothesis in order to check if the mean enrollment at four-year colleges is higher than at two-year colleges in the United States , the system of hypothesis would be:

Null hypothesis: $\mu_{1}-\mu_{2}\leq 0$

Alternative hypothesis: $\mu_{1} - \mu_{2}> 0$

We can assume that the normal distribution is assumed since we have a large sample size for each case n>30. So then the sample mean can be assumed as normally distributed.

Part 1: The statistic

$t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}$ (1)

And the degrees of freedom are given by $df=n_1 +n_2 -2=35+35-2=68$

Replacing we got

$t=\frac{(5135-4436)-0}{\sqrt{\frac{783^2}{35}+\frac{553^2}{35}}}}=4.31$

Part 2: P value

Since is a right tailed test the p value would be:

$p_v =P(t_{68}>4.31)=0.000022$

Comparing the p value we see that is lower compared to the significance level of 0.01 so then we can reject the null hypothesis and we can conclude that the mean for the four year college is significantly higher than the mean for the two year college and then the claim makes sense

6 0

3 years ago

What is the absolute value of |5+8a|=53?

kozerog [31]

Answer:

a = 6 or a = -58/8

Step-by-step explanation:

I think you meant to ask us to solve for a.

There are two cases to solve for:

5 + 8a = 53 or -(5 + 8a) = 53

8a = 48 or 5 + 8a = -53

a = 6 or 8a = -58

a = 6 or a = -58/8

7 0

3 years ago

HELP FAST! 15 POINTS! PLEASE HELP!

masha68 [24]

Answer:

Step-by-step explanation:

Basically, in each option, you have to find the relationship between x and y (for one pair). Like in option #1 - how do you get 1.5 from 3? divide it by 2. Now you apply it for all numbers. It the outcome the same? No. 7 ÷ 2 ≠ 3.

You do the same process for all, until you are left with the last one. y = 50x

The last option is the answer.

6 0

3 years ago

WILL NAME BRAINLEST ONGG <br> PLEASE HELP

Leona [35]

Answer:

since angle 1 and angle 2 are equal both angles can only be 90.

and given that b is perpendicular to d so angle at point of interception of bd is also 90.

it is also given that c and d are parallel so both line have same angle on point of interception of line a and b and angle formed is also the same on line a and b .....so a and b are parallell

Step-by-step explanation:

8 0

3 years ago