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Citrus2011 [14]

3 years ago

7

Find the volume of this prism.

2 answers:

s2008m [1.1K]3 years ago

8 0

It is D because you have to multiply 8x7 which is 56 and then multiply 14x 56 which is the answer 784.

enyata [817]3 years ago

7 0

Isnt it letter D ? Not sure tho

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The probability that a lab specimen contains high levels of contamination is 0.10. Five samples are checked, and the samples are

raketka [301]

Answer:

(a) 0.59049 (b) 0.32805 (c) 0.40951

Step-by-step explanation:

Let's define

$A_{i}$ : the lab specimen number i contains high levels of contamination for i = 1, 2, 3, 4, 5, so,

$P(A_{i})=0.1$ for i = 1, 2, 3, 4, 5

The complement for $A_{i}$ is given by

$A_{i}^{$c$}$ : the lab specimen number i does not contains high levels of contamination for i = 1, 2, 3, 4, 5, so

P( $A_{i}^{$c$}$ )=0.9 for i = 1, 2, 3, 4, 5

(a) The probability that none contain high levels of contamination is given by

P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )= $(0.9)^{5}$ =0.59049 because we have independent events.

(b) The probability that exactly one contains high levels of contamination is given by

P( $A_{1}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}$ )=5×(0.1)× $(0.9)^{4}$ =0.32805

because we have independent events.

(c) The probability that at least one contains high levels of contamination is

P( $A_{1}$ ∪ $A_{2}$ ∪ $A_{3}$ ∪ $A_{4}$ ∪ $A_{5}$ )=1-P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )=1-0.59049=0.40951

6 0

2 years ago

Researchers conducted a study to find out if there is a difference in the use of ereaders by different age groups. randomly sele

ikadub [295]

No, there is significant difference in the use of e readers by different age groups.

Given sample 1 ( 29 years old) $n_{1}$ =628, $p_{1}$ =7%, sample 2( 30 years old) $n_{2}$ =2309, $p_{2}$ =0.11.

We have to first form hypothesis one null hypothesis and other alternate hypothesis.

$H_{0}:$ π1-π2=0

$H_{1}:$ π1-π2≠0

α=0.05

Difference between proportions $p_{1}-p_{2} =-0.04$

$p_{d}=0.07-0.11=-0.04$

The pooled proportion needed to calculate standard error is:

$p=(X_{1} -X_{2} )/(n_{1} +n_{2} )$

=(44+254)/(628+2309)

=0.10146

The estimated standard error of difference between means is computed using the formula:

$S_{p_{1} -p_{2} }=\sqrt{p(1-p)/m_{1}+p(1-p)/n_{2} }$

= $\sqrt{0.101*0.899/628+0.101*0.899/2309}$

= $\sqrt{0.000143+0.00003}$

= $\sqrt{0.000173}$

=0.01315

Z= Pd-(π1-π2)/ $S_{p_{1} -p_{2} }$

=-0.04-0/0.013

=-3.0769

This test is a two tailed test so the p value for this test is calculated as (using z table)

p value:2 P(Z<-3.0769)

=2*0.002092

=0.004189

P value< significance level of 5%.

Hence there is enough evidence to show the claim that there is a significant difference in the use of e readers by different age groups.

Learn more about hypothesis at brainly.com/question/11555274

#SPJ4

Question is incomplete as it also includes:

Significance level of 5%.

4 0

2 years ago

The inequality x ≥ -3 is satisfied by point (0, -2).<br> True<br> False

denpristay [2]

This inequality is true doll!!

8 0

3 years ago

Read 2 more answers

100 pts Eight stuffed animals need to be arranged in a line. There is one bear and one zebra. What is the exact probability that

shtirl [24]

Answer:

1/56

Step-by-step explanation:

P( bear is first) = bear/ total = 1/8

Then there are 7 animals left

P( zebra is 2nd) = zebra /total = 1/7

P( bear, then zebra) = 1/8 * 1/7 = 1/56

3 0

3 years ago

Read 2 more answers

In a sport competition Alyssa was penalized -16 points in each of the 4 events how many points was she penalized in each event?

borishaifa [10]

-4 because -16-4=-4. Negative and a positive make a negative

5 0

3 years ago