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Likurg_2 [28]
3 years ago
8

Pamela Is 10 years older than Jiri. The sum of their ages is 96. What is Jiri’s age

Mathematics
2 answers:
Solnce55 [7]3 years ago
8 0
43 is jiri and 53 is pamela
givi [52]3 years ago
6 0

Answer:

86

Step-by-step explanation:

The answer is 86 because, 96-10= 86... Thats how u find Jiri's age... Hope this helps!

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Camille had a very fun birthday party with lots of friends and family attending. The party lasted for 3 hours. She and her frien
Elena-2011 [213]

Camille spent 22 minutes and 30 seconds opening presents.

Since Camille had a very fun birthday party with lots of friends and family attending, and the party lasted for 3 hours, and she and her friends played games for 3/8 of the time, ate pizza and cake for 50% of the time , and spent the remainder of the time opening presents, to determine the amount of time spent opening presents the following calculation must be performed, posing the following linear equation:

Total time - time spent on other activities = time spent opening presents

  • 3 - (3 x 3/8) - (3 x 0.5) = X
  • 3 - 1.125 - 1.5 = X
  • 3 - 2.625 = X
  • 0.375 = X

  • 1 = 60
  • 0.375 = X
  • 0.375 x 60 = X
  • 22.5 = X

Therefore, Camille spent 22 minutes and 30 seconds opening presents.

Learn more in brainly.com/question/11897796

7 0
2 years ago
Is 6.56556555 a rational or irrational number
lord [1]

Answer:

yes

Step-by-step explanation:

if it is a terminating decimal(a decimal that ends) it will be rational

3 0
3 years ago
Read 2 more answers
If popsicle were black would there be white?
bija089 [108]
Yes, of course there'd be white
6 0
3 years ago
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A scientist wants to find the average weight of fish living in a large pond. So she will use the weights of 25 fish from the pon
lidiya [134]

Answer:

Use interval estimation to compute the average weight of fish living in a large pond.

Step-by-step explanation:

In statistic, point estimation comprises of the use of sample data to estimate a distinct data value (known as a point estimate) which is to function as a "best guess" or "best estimate" of an unidentified population parameter.  

The point estimate of the population mean (<em>µ</em>) is the sample mean (\bar x).

Now according to the law of large numbers, in probability concept, states that as we increase the sample size, the mean of the sample (\bar x) approaches the whole population mean (µ).

That is, \bar x →  <em>µ</em> as <em>n</em> → ∞.

The scientist uses the weights of 25 fish from the pond to find the average weight.

The sample is large, but not large enough.

So, use interval estimation to compute the average weight of fish living in a large pond.

The interval estimation is the best way to estimate the true value of a parameter. This is because the interval of values computed for the parameter has certain probability of containing the true parameter value.

The (1 - <em>α</em>)% confidence interval for population mean is an interval estimate.

4 0
3 years ago
Somebody please assist me here
Anettt [7]

The base case of n=1 is trivially true, since

\displaystyle P\left(\bigcup_{i=1}^1 E_i\right) = P(E_1) = \sum_{i=1}^1 P(E_i)

but I think the case of n=2 may be a bit more convincing in this role. We have by the inclusion/exclusion principle

\displaystyle P\left(\bigcup_{i=1}^2 E_i\right) = P(E_1 \cup E_2) \\\\ P\left(\bigcup_{i=1}^2 E_i\right) = P(E_1) + P(E_2) - P(E_1 \cap E_2) \\\\ P\left(\bigcup_{i=1}^2 E_i\right) \le P(E_1) + P(E_2) \\\\ P\left(\bigcup_{i=1}^2 E_i\right) \le \sum_{i=1}^2 P(E_i)

with equality if E_1\cap E_2=\emptyset.

Now assume the case of n=k is true, that

\displaystyle P\left(\bigcup_{i=1}^k E_i\right) \le \sum_{i=1}^k P(E_i)

We want to use this to prove the claim for n=k+1, that

\displaystyle P\left(\bigcup_{i=1}^{k+1} E_i\right) \le \sum_{i=1}^{k+1} P(E_i)

The I/EP tells us

\displaystyle P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) = P\left(\left(\bigcup\limits_{i=1}^k E_i\right) \cup E_{k+1}\right) \\\\ P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) = P\left(\bigcup\limits_{i=1}^k E_i\right) + P(E_{k+1}) - P\left(\left(\bigcup\limits_{i=1}^k E_i\right) \cap E_{k+1}\right)

and by the same argument as in the n=2 case, this leads to

\displaystyle P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) = P\left(\bigcup\limits_{i=1}^k E_i\right) + P(E_{k+1}) - P\left(\left(\bigcup\limits_{i=1}^k E_i\right) \cap E_{k+1}\right) \\\\ P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) \le P\left(\bigcup\limits_{i=1}^k E_i\right) + P(E_{k+1})

By the induction hypothesis, we have an upper bound for the probability of the union of the E_1 through E_k. The result follows.

\displaystyle P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) \le P\left(\bigcup\limits_{i=1}^k E_i\right) + P(E_{k+1}) \\\\ P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) \le \sum_{i=1}^k P(E_i) + P(E_{k+1}) \\\\ P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) \le \sum_{i=1}^{k+1} P(E_i)

5 0
2 years ago
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