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Nezavi [6.7K]
3 years ago
9

Your math Teacher has assigned homework in 9/10 of your last classes. Describe the likelihood of having homework tonight.

Mathematics
1 answer:
Solnce55 [7]3 years ago
4 0

Answer:

It is very likely that you will have homework tonight. Have fun! :)

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For some postive value of Z, the probability that a standardized normal variable is between 0 and Z is 0.3770. The value of Z is
Talja [164]

Answer:

1.16

Step-by-step explanation:

Given that;

For some positive value of Z, the probability that a standardized normal variable is between 0 and Z is 0.3770.

This implies that:

P(0<Z<z) = 0.3770

P(Z < z)-P(Z < 0) = 0.3770

P(Z < z) = 0.3770 + P(Z < 0)

From the standard normal tables , P(Z < 0)  =0.5

P(Z < z) = 0.3770 + 0.5

P(Z < z) =  0.877

SO to determine the value of z for which it is equal to 0.877, we look at the

table of standard normal distribution and locate the probability value of 0.8770. we advance to the  left until the first column is reached, we see that the value was 1.1.  similarly, we did the same in the  upward direction until the top row is reached, the value was 0.06.  The intersection of the row and column values gives the area to the two tail of z.   (i.e 1.1 + 0.06 =1.16)

therefore, P(Z ≤ 1.16 ) = 0.877

8 0
3 years ago
Triangle congruence worksheet
lesya [120]

Answer:

NC

Step-by-step explanation:

We know it is not SSS or AAS or ASA

On one triangle we have SAS but we don't see the same congruent markings so we cannot tell.

4 0
3 years ago
the half-life of chromium-51 is 38 days. If the sample contained 510 grams. How much would remain after 1 year?​
madam [21]

Answer:

About 0.6548 grams will be remaining.  

Step-by-step explanation:

We can write an exponential function to model the situation. The standard exponential function is:

f(t)=a(r)^t

The original sample contained 510 grams. So, a = 510.

Each half-life, the amount decreases by half. So, r = 1/2.

For t, since one half-life occurs every 38 days, we can substitute t/38 for t, where t is the time in days.

Therefore, our function is:

\displaystyle f(t)=510\Big(\frac{1}{2}\Big)^{t/38}

One year has 365 days.

Therefore, the amount remaining after one year will be:

\displaystyle f(365)=510\Big(\frac{1}{2}\Big)^{365/38}\approx0.6548

About 0.6548 grams will be remaining.  

Alternatively, we can use the standard exponential growth/decay function modeled by:

f(t)=Ce^{kt}

The starting sample is 510. So, C = 510.

After one half-life (38 days), the remaining amount will be 255. Therefore:

255=510e^{38k}

Solving for k:

\displaystyle \frac{1}{2}=e^{38k}\Rightarrow k=\frac{1}{38}\ln\Big(\frac{1}{2}\Big)

Thus, our function is:

f(t)=510e^{t\ln(.5)/38}

Then after one year or 365 days, the amount remaining will be about:

f(365)=510e^{365\ln(.5)/38}\approx 0.6548

5 0
3 years ago
Someone help me with this problem.
Klio2033 [76]
Download Photomath and there
8 0
3 years ago
Help plz it due today
Angelina_Jolie [31]

Answer:

Point A  - 1/2 and - 1/4

Point B  - 2 1/2 and 1 1/2

Point C 1/4 and 1/2

Step-by-step explanation:

6 0
3 years ago
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