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2 years ago

PLZ DO BOTH PARTS FOR 100 POINTS AND ILL DO ANYTHING!!!!

Mathematics

2 answers:

Kitty [74]2 years ago

8 0

Step-by-step explanation:

<u>Multiply the energy produced in 1 second by the number of seconds:</u>

$3.9*10^{33} * 1.55*10^7 = 6.045*10^{40}$ erg

10⁻³ is same as 0.001 times and 10³ is same as 1000 times.

1.6*1000 = 1600 mm = 160 cm = 1.6 m is way too big for the measurement of the distance between the tracks on a CD.

So the first option is correct.

Aleonysh [2.5K]2 years ago

7 0

Multiply the energy produced in 1 second by the number of seconds:

Part A

3.9*10^{33} * 1.55*10^7 = 6.045*10^{40}

Part B

10⁻³ is same as 0.001 times and 10³ is same as 1000 times.

1.6*1000 = 1600 mm = 160 cm = 1.6 m is way too big for the measurement of the distance between the tracks on a CD.

So the first option is correct.

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OLga [1]

Answer:

n is less than or equal to: 10

Step-by-step explanation:

10n+20=120

subtract 20

10n=100

divide by 10

n=10

7 0

3 years ago

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The ratio of the sales tax to the amount of a purchase is a fixed number in Town Q. The table shows the sales tax for a purchase

Volgvan

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3 years ago

Discrete R.V. Assume a student shows up for EGR 280 completely unprepared and the instructor gives a pop quiz. Assume there are

Tom [10]

Answer:

a) p=0.2

b) probability of passing is 0.01696

c) The expected value of correct questions is 1.2

Step-by-step explanation:

a) Since each question has 5 options, all of them equally likely, and only one correct answer, then the probability of having a correct answer is 1/5 = 0.2.

b) Let X be the number of correct answers. We will model this situation by considering X as a binomial random variable with a success probability of p=0.2 and having n=6 samples. We have the following for k=0,1,2,3,4,5,6

$P(X=k) = \binom{n}{k}p^{k}(1-p)^{n-k} = \binom{6}{k}0.2^{k}(0.8)^{6-k}$ .

Recall that $\binom{n}{k}= \frac{n!}{k!(n-k)!}$ In this case, the student passes if X is at least four correct questions, then

$P(X\geq 4) = P(X=4)+P(X=5)+P(X=6)=\binom{6}{4}0.2^{4}(0.8)^{6-4}+\binom{6}{5}0.2^{5}(0.8)^{6-5}+\binom{6}{6}0.2^{6}(0.8)^{6-6}= 0.01696$

c)The expected value of a binomial random variable with parameters n and p is $E[X] = np$ . IN our case, n=6 and p =0.2. Then the expected value of correct answers is $6\cdot 0.2 = 1.2$