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maksim [4K]
3 years ago
15

Answer ASAP Find the area for this shape

Mathematics
2 answers:
antiseptic1488 [7]3 years ago
8 0
A = \frac{a+b}{2} (h)
A = \frac{3.8+12}{2} (4.6)
A ≈ 36.34

The area is approximately 36.34 cm²
expeople1 [14]3 years ago
5 0
A = 1/2(3.8 + 12)(4.6)
A = 36.34

answer
36.34 cm^2
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PLS HELP ME ASAL FOR 7, 8, and 9. A and B! (SHOW WORK!!!!) + LOTS IF POINTS!! (BE SPECIFIC ON WHICH your ANSWERING)
nevsk [136]
Ok so
Q.1=B
Q.2=B
Q.3(a)=(d-45)+m+65
      (b)=d+m+20  
5 0
3 years ago
Please ΔEFU ~ ΔVWU. Find the length of UF.
koban [17]
<h3>Answer: UF = 11</h3>

=======================================================

Explanation:

In EFU, we see that EU are the first and last letters. In VWU, we see that VU are the first and last letters. This order is important to be able to pair EU with VU

So EU/VU is one ratio

The other ratio is UF and UW, which are the last two letters of EFU and VWU respectively. So the other ratio is UF/UW

Set the ratios equal to each other and solve for x

EU/VU = UF/UW

7/35 = x/55

7*55 = 35*x ... cross multiply

385 = 35x

35x = 385

x = 385/35 .... divide both sides by 35

x = 11

UF = 11

------------

Or you could note how UV = 35 is exactly 5 times larger compared to EU = 7

That must mean WU = 55 must also be 5 times larger compared to UF = x

So,

WU = 5*UF

55 = 5x

5x = 55

x = 55/5

x = 11

UF = 11

This is similar to the previous section because UV/EU = 35/7 = 5 is the scale factor.

4 0
4 years ago
In a particular faculty 60% of students are men and 40% are women. In a random sample of 50 students what is the probability tha
zimovet [89]

Answer:

a) The expected value is given by:

E(X) = np = 50*0.4 = 20

and the variance is given by:

Var(X) =np(1-p) = 50*0.4*(1-0.4) = 12

b) P(X>25)= 1-P(X\leq 25)

And we can find this probability with the following Excel code:

=1-BINOM.DIST(25,50,0.4,TRUE)

And we got:

P(X>25)= 1-P(X\leq 25)=0.0573

c) 1) Random sample (assumed)

2) np= 50*0.4= 20 >10

n(1-p) =50*0.6= 30>10

3) Independence (assumed)

Since the 3 conditions are satisfied we can use the normal approximation:

X \sim N(\mu = 20 , \sigma= 3.464)

d) P(X>25) = 1-P(Z< \frac{25-20}{3.464}) = 1-P(z

e) P(X>25)= P(X>25.5) = 1-P(X \leq 25.5)

P(X>25)= P(X>25.5) = 1-P(X \leq 25.5)= 1-P(Z

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Let X the random variable of interest, on this case we now that:

X \sim Binom(n=50, p=0.4)

The probability mass function for the Binomial distribution is given as:

P(X)=(nCx)(p)^x (1-p)^{n-x}

Where (nCx) means combinatory and it's given by this formula:

nCx=\frac{n!}{(n-x)! x!}

Part a

The expected value is given by:

E(X) = np = 50*0.4 = 20

and the variance is given by:

Var(X) =np(1-p) = 50*0.4*(1-0.4) = 12

Part b

For this case we want to find this probability:

P(X>25)= 1-P(X\leq 25)

And we can find this probability with the following Excel code:

=1-BINOM.DIST(25,50,0.4,TRUE)

And we got:

P(X>25)= 1-P(X\leq 25)=0.0573

Part c

1) Random sample (assumed)

2) np= 50*0.4= 20 >10

n(1-p) =50*0.6= 30>10

3) Independence (assumed)

Since the 3 conditions are satisfied we can use the normal approximation:

X \sim N(\mu = 20 , \sigma= 3.464)

Part d

We want this probability:

P(X>25) = 1-P(Z< \frac{25-20}{3.464}) = 1-P(z

Part e

For this case we use the continuity correction and we have this:

P(X>25)= P(X>25.5) = 1-P(X \leq 25.5)

P(X>25)= P(X>25.5) = 1-P(X \leq 25.5)= 1-P(Z

4 0
4 years ago
I need help. i will give brainiest as soon as possible
SCORPION-xisa [38]

Answer:

B

Step-by-step explanation:

Let me know if you need an explanation.

3 0
3 years ago
Read 2 more answers
Lawrence performs a survey to determine the average number of minutes of exercise seventh-grade athletes get in one week. He get
RoseWind [281]

Answer:

1) Is more representative

Step-by-step explanation:

The problem with his selection is that maybe there are few students participating in certain sport and those students maybe do quite more excercise than the rest (or quite less). This will modify the results because the sample he selected is biased. This problem wont be solved by method 3 or 4, because he is still selecting students that may modify heavily the results with a high probability

This problem will also appear if he choose a sample by class. Maybe, in a class there are quite few students, and selecting from class will make those students appear quite more often than, lets say, a 7th grade student selected at random, therefore the selection is biased in this case as well.

If he has a list with all seventh grade students, each student is equallly likely to be selected and as a consequence, the the results wont be biased. Approach 1 is the best one.

5 0
3 years ago
Read 2 more answers
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