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stiks02 [169]
3 years ago
8

Point P is the midpoint of both MR and NQ, which are on the same line. If MN=26, NP=3x+4, and QR=6x+2, find NP.

Mathematics
1 answer:
sammy [17]3 years ago
7 0

Answer: NP= 16 cm

Explanation:

As we shown in the figure below :

Since P is the midpoint of MN and  NQ

So, We consider ΔMNP and ΔPQR,

∠MPN=∠QPR  (∵ Vertically opposite angles are equal)

MP=PR  (∵ P is the mid point of MR)

NP=PQ  (∵P is the midpoint of NQ)

So, ΔMNQ ≅ ΔPQR  (∵By SAS congruence )

∴ 26=6x+2 \text{(By CPCT )}\\26-2=6x\\24=6x\\\frac{24}{6}=x\\4=x

So, NP=3x+4=3\times 4+4=12+4=16 cm

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Step-by-step explanation:

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3 years ago
Select the simplification that accurately explains the following statement.
lutik1710 [3]

Answer:

Option (b) is correct.

(2^\frac{1}{4} )^4=2^\frac{1}{4} \times 2^\frac{1}{4}\times 2^\frac{1}{4}\times 2^\frac{1}{4}=2^{(\frac{1}{4}+ \frac{1}{4}+ \frac{1}{4}+ \frac{1}{4} )}=2^{1}=2

Step-by-step explanation:

Given: (2^\frac{1}{4} )^4

We have too choose the correct simplification for the given statement.

Consider (2^\frac{1}{4} )^4

Using property of exponents, (a^m)^n=a^m\times a^m\times a^m\times ....\times (n\ times)

We have,

(2^\frac{1}{4} )^4=2^\frac{1}{4} \times 2^\frac{1}{4}\times 2^\frac{1}{4}\times 2^\frac{1}{4}

Again applying property of exponents  a^m\times a^m=a^{n+m}

We have,

(2^\frac{1}{4} )^4=2^{(\frac{1}{4}+ \frac{1}{4}+ \frac{1}{4}+ \frac{1}{4} )}

Simplify, we have,

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finlep [7]

Answer:

a) There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

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c) There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

Step-by-step explanation:

There are only two possible outcomes. Either the student use credit cards because of the rewards program, or they use for other reason. So, we can solve this problem by the binomial distribution.

Binomial probability

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

In which C_{n,x} is the number of different combinatios of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And \pi is the probability of X happening.

In this problem, we have that:

10 student are sampled, so n = 10

34% of college students say they use credit cards because of the rewards program, so \pi = 0.34

(a) exactly​ two

This is P(X = 2).

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873

There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

(b) more than​ two

This is P(X > 2).

Either a value is larger than two, or it is smaller of equal. The sum of the decimal probabilities must be 1. So:

P(X \leq 2) + P(X > 2) = 1

P(X > 2) = 1 - P(X \leq 2)

In which

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)

So

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

P(X = 0) = C_{10,0}.(0.34)^{0}.(0.66)^{10} = 0.0157

P(X = 1) = C_{10,1}.(0.34)^{1}.(0.66)^{9} = 0.0808

P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873

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(c) between two and five inclusive

This is:

P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873

P(X = 3) = C_{10,3}.(0.34)^{3}.(0.66)^{7} = 0.2573

P(X = 4) = C_{10,4}.(0.34)^{4}.(0.66)^{6} = 0.2320

P(X = 5) = C_{10,5}.(0.34)^{5}.(0.66)^{5} = 0.1434

P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.1873 + 0.2573 + 0.2320 + 0.1434 = 0.82

There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

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Answer:

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