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2 years ago

One number is 2 less than a second number. Twice the second number is 5 less than 3 times the first. Find the two numbers.

Mathematics

1 answer:

Evgen [1.6K]2 years ago

8 0

Answer:

9 and 11

Step-by-step explanation:

First number = x -2

Second number = x

and 2x = 3(x-2) - 5

2x = 3x - 6 - 5

x = 11

first number = 11-2 = 9

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Step-by-step explanation:

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3 years ago

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A prticular type of tennis racket comes in a midsize versionand an oversize version. sixty percent of all customers at acertain

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Answer:

a) P(x≥6)=0.633

b) P(4≤x≤8)=0.8989 (one standard deviation from the mean).

c) P(x≤7)=0.8328

Step-by-step explanation:

a) We can model this a binomial experiment. The probability of success p is the proportion of customers that prefer the oversize version (p=0.60).

The number of trials is n=10, as they select 10 randomly customers.

We have to calculate the probability that at least 6 out of 10 prefer the oversize version.

This can be calculated using the binomial expression:

$P(x\geq6)=\sum_{k=6}^{10}P(k)=P(6)+P(7)+P(8)+P(9)+P(10)\\\\\\P(x=6) = \binom{10}{6} p^{6}q^{4}=210*0.0467*0.0256=0.2508\\\\P(x=7) = \binom{10}{7} p^{7}q^{3}=120*0.028*0.064=0.215\\\\P(x=8) = \binom{10}{8} p^{8}q^{2}=45*0.0168*0.16=0.1209\\\\P(x=9) = \binom{10}{9} p^{9}q^{1}=10*0.0101*0.4=0.0403\\\\P(x=10) = \binom{10}{10} p^{10}q^{0}=1*0.006*1=0.006\\\\\\P(x\geq6)=0.2508+0.215+0.1209+0.0403+0.006=0.633$

b) We first have to calculate the standard deviation from the mean of the binomial distribution. This is expressed as:

$\sigma=\sqrt{np(1-p)}=\sqrt{10*0.6*0.4}=\sqrt{2.4}=1.55$

The mean of this distribution is:

$\mu=np=10*0.6=6$

As this is a discrete distribution, we have to use integer values for the random variable. We will approximate both values for the bound of the interval.

$LL=\mu-\sigma=6-1.55=4.45\approx4\\\\UL=\mu+\sigma=6+1.55=7.55\approx8$

The probability of having between 4 and 8 customers choosing the oversize version is:

$P(4\leq x\leq 8)=\sum_{k=4}^8P(k)=P(4)+P(5)+P(6)+P(7)+P(8)\\\\\\P(x=4) = \binom{10}{4} p^{4}q^{6}=210*0.1296*0.0041=0.1115\\\\P(x=5) = \binom{10}{5} p^{5}q^{5}=252*0.0778*0.0102=0.2007\\\\P(x=6) = \binom{10}{6} p^{6}q^{4}=210*0.0467*0.0256=0.2508\\\\P(x=7) = \binom{10}{7} p^{7}q^{3}=120*0.028*0.064=0.215\\\\P(x=8) = \binom{10}{8} p^{8}q^{2}=45*0.0168*0.16=0.1209\\\\\\P(4\leq x\leq 8)=0.1115+0.2007+0.2508+0.215+0.1209=0.8989$

c. The probability that all of the next ten customers who want this racket can get the version they want from current stock means that at most 7 customers pick the oversize version.

Then, we have to calculate P(x≤7). We will, for simplicity, calculate this probability substracting P(x>7) from 1.

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Answer:

The Proof and Explanation for

Part C ,

Qs 9 and

Qs 10 are below.

Step-by-step explanation:

PART C .

Given:

AD || BC ,

To Prove:

ΔAED ≅ ΔCEB

Proof:

Statement Reason

1. AD || BC 1. Given

2. ∠A ≅ ∠C 2. Alternate Angles Theorem as AD || BC

3. ∠AED ≅ ∠CEB 3. Vertical Opposite Angle Theorem.

4. AE ≅ EC 4. Given

5. ΔAED ≅ ΔCEB 5. By A-S-A congruence test....Proved

Qs 9)

Given:

AB ≅ BC ,

∠ABD ≅ ∠CBD

To Prove:

∠A ≅ ∠C

Proof:

Statement Reason

1. ∠ABD ≅ ∠CBD 1. Given

2. AB ≅ CB 2. Given

3. BD ≅ BD 3. Reflexive Property

4. ΔABD ≅ ΔCBD 4. By S-A-S congruence test

5. ∠A ≅ ∠C 5. Corresponding parts of congruent Triangles Proved.

Qs 10)

Given:

∠MCI ≅ ∠AIC

MC ≅ AI

To Prove:

ΔMCI ≅ ΔAIC

Proof:

Statement Reason

1. ∠MCI ≅ ∠AIC 1. Given

2. MC ≅ AI 2. Given

3. CI ≅ CI 3. Reflexive Property

4. ΔMCI ≅ ΔAIC 4. By S-A-S congruence test

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3 years ago