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svetlana [45]
3 years ago
10

Math question 2! Thanks if you help

Mathematics
2 answers:
ruslelena [56]3 years ago
8 0
The answer to this question is 15.6m
Greeley [361]3 years ago
6 0

Answer: 15.6 m

Step-by-step explanation:

Two legs of right angle in right triangle given are 12m and 10m.

For the third side use Pythagoras theorem:

C^2= a^2+b^2

C=√[(12^2)+(10^2)]

C=√[144+100]

C=√[244]= 15.6 m

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x^2 + (7)x +18

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What are 5 numbers that have 5, 11, and 2 as prime factors
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110, 220, 330, 440, 550

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coin $a$ is flipped three times and coin $b$ is flipped four times. what is the probability that the number of heads obtained fr
ivolga24 [154]

The probability that the number of heads obtained from flipping the two fair coins is the same is 35/128.

Probability:

Probability means the fraction of favorable outcome and the total number of outcomes.

So it can be written as,

Probability = Favorable outcomes / Total outcomes

Given,

The coin a is flipped three times and coin b is flipped four times.

Here we need to find the probability that the number of heads obtained from flipping the two fair coins is the same.

We know that,

There are 4 ways that the same number of heads will be obtained;

0, 1, 2, or 3 heads.

The probability of both getting 0 heads is

$\left(\frac12\right)^3{3\choose0}\left(\frac12\right)^4{4\choose0}=\frac1{128}$

Probability of getting 1 head,

$\left(\frac12\right)^3{3\choose1}\left(\frac12\right)^4{4\choose1}=\frac{12}{128}$

Probability of getting 2 heads is,

$\left(\frac12\right)^3{3\choose2}\left(\frac12\right)^4{4\choose2}=\frac{18}{128}$

And the probability of getting 3 heads is,

$\left(\frac12\right)^3{3\choose3}\left(\frac12\right)^4{4\choose3}=\frac{4}{128}$

Therefore, the probability that the number of heads obtained from flipping the two fair coins is the same is,

=> (1/128) + (12/128) + (18/128) + (4/128)

=> 35/128.

To know more about probability here

brainly.com/question/14210034

#SPJ4

4 0
1 year ago
The following data represent the monthly phone use, in minutes, of a customer enrolled in a fraud prevention program for the pas
Illusion [34]

Answer:

The answer is "717.25 minutes".

Step-by-step explanation:

In this question first, we arrange the value in ascending order that are:

307,311,321,322,354,363,377,406,425,435,461,464,474,499,505,513,517,534,548

The median is always in an orderly position that is = \frac{(n+1)}{2}.

\to \frac{(n+1)}{2} = \frac{(20+1)}{2} = 10.5  position orders

10^{th} \ and \ 11^{th} place an average of observations so, the

Average = \frac{(425 +435)}{2} = \frac{(860)}{2} =430

Because as medium stands at 10.5 that median is as below 10 are greater than the level.

Q_1 often falls throughout the ordered BELOW average is = \frac{(n+1)}{2} place.

\to \frac{(n+1)}{2} = \frac{(10+1)}{2} = 5.5^{th}ordered position

Average 5^{th} \ and \ 6^{th}  location findings

Q_1 = \frac{(354+363)}{2} = \frac{(717)}{2}= 358.5

In  = \frac{(n+1)}{2} the ordered place, Q3 always falls ABOVE the median.

\to \frac{(n+1)}{2} = \frac{(10+1)}{2} = 5.5^{th} ordered At median ABOVE 

Consequently Q_3 falls between 5^{th} \ and \ 6^{th} ABOVE the median position

15^{th}\  and \ 16^{th}place average of observations

\to Q_3 = \frac{(499+505)}{2}=\frac{(1004)}{2}  = 502

\to IQR = Q_3 - Q_1= 502-358.5= 143.5  

\to \text{Upper fence} = Q_3 + 1.5 \times IQR= 502 + 1.5 \times 143.5 = 717.25

4 0
3 years ago
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