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Margaret [11]
3 years ago
5

8×25×23 strategies 4 grade

Mathematics
1 answer:
DerKrebs [107]3 years ago
4 0
8×25=200
and 
200×23=4600

so 8×25×23=4600
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The amount $180.00 is what percent greater than $135.00? 
sergiy2304 [10]

Answer:

Its Option C

Step-by-step explanation:

The difference is 180 - 135 = $45

So  it is 45 * 100 / 135  

= 33.33% greater

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Which expression is equivalent to 2(4 + x) + 3 ?
Elden [556K]
The correct answer is A. 2*4+x+3
6 0
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Bryce has 28 coins in his pocket, all of which are dimes and quarters. if the total value of his change is 445 cents, how many d
raketka [301]
He has 11 quarters and 17 dimes.
11 x .25 = 2.75       17 x .1 = 1.7        2.75 + 1.7 = 4.45
Hope this helps
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7. Consider the diagram below. Please find the value of w.
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Read 2 more answers
32​% of college students say they use credit cards because of the rewards program. You randomly select 10 college students and a
Tems11 [23]

Answer:

a) P(X=2)=(10C2)(0.32)^2 (1-0.32)^{10-2}=0.211

b) P(X> 2)=1-P(X\leq 2)=1-[0.0211+0.0995+0.211]=0.668

c) P(2 \leq x \leq 5)=0.211+0.264+0.218+0.123=0.816

Step-by-step explanation:

1) 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".

2) Solution to the problem

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

X \sim Binom(n=10, p=0.32)

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

P(X=2)=(10C2)(0.32)^2 (1-0.32)^{10-2}=0.211

Part b

P(X> 2)=1-P(X\leq 2)=1-[P(X=0)+P(X=1)+P(X=2)]

P(X=0)=(10C0)(0.32)^0 (1-0.32)^{10-0}=0.0211  

P(X=1)=(10C1)(0.32)^1 (1-0.32)^{10-1}=0.0995

P(X=2)=(10C2)(0.32)^2 (1-0.32)^{10-2}=0.211

P(X> 2)=1-P(X\leq 2)=1-[0.0211+0.0995+0.211]=0.668

Part c

P(2 \leq x \leq 5)=P(X=2)+P(X=3)+P(X=4)+P(X=5)

P(X=2)=(10C2)(0.32)^2 (1-0.32)^{10-2}=0.211

P(X=3)=(10C3)(0.32)^3 (1-0.32)^{10-3}=0.264

P(X=4)=(10C4)(0.32)^4 (1-0.32)^{10-4}=0.218

P(X=5)=(10C5)(0.32)^5 (1-0.32)^{10-5}=0.123

P(2 \leq x \leq 5)=0.211+0.264+0.218+0.123=0.816

6 0
3 years ago
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