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3 years ago

Solve 2x+8<10 or 2x+8>20

Mathematics

1 answer:

galben [10]3 years ago

8 0

Answer:

1 Add 88 to both sides.

2x=10+8

2 Simplify 10+810+8 to 1818.

2x=18

3 Divide both sides by 22.

x=\frac{18}{2}

4 Simplify \frac{18}{2}

to 99.

x=9

Step-by-step explanation:

2x−8=10

1 Let x=9x=9.

2\times 9-8=10

2×9−8=10

2 Simplify 2\times 92×9 to 1818.

18-8=10

18−8=10

3 Simplify 18-818−8 to 1010.

10=10

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4 years ago

The price of a watch was increased by 10% to £132. What was the price before the increase?

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3 years ago

Read 2 more answers

Let a1, a2, a3, ... be a sequence of positive integers in arithmetic progression with common difference

Bezzdna [24]

Since $a_1,a_2,a_3,\cdots$ are in arithmetic progression,

$a_2 = a_1 + 2$

$a_3 = a_2 + 2 = a_1 + 2\cdot2$

$a_4 = a_3+2 = a_1+3\cdot2$

$\cdots \implies a_n = a_1 + 2(n-1)$

and since $b_1,b_2,b_3,\cdots$ are in geometric progression,

$b_2 = 2b_1$

$b_3=2b_2 = 2^2 b_1$

$b_4=2b_3=2^3b_1$

$\cdots\implies b_n=2^{n-1}b_1$

Recall that

$\displaystyle \sum_{k=1}^n 1 = \underbrace{1+1+1+\cdots+1}_{n\,\rm times} = n$

$\displaystyle \sum_{k=1}^n k = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}2$

It follows that

$a_1 + a_2 + \cdots + a_n = \displaystyle \sum_{k=1}^n (a_1 + 2(k-1)) \\\\ ~~~~~~~~ = a_1 \sum_{k=1}^n 1 + 2 \sum_{k=1}^n (k-1) \\\\ ~~~~~~~~ = a_1 n + n(n-1)$

so the left side is

$2(a_1+a_2+\cdots+a_n) = 2c n + 2n(n-1) = 2n^2 + 2(c-1)n$

Also recall that

$\displaystyle \sum_{k=1}^n ar^{k-1} = \frac{a(1-r^n)}{1-r}$

so that the right side is

$b_1 + b_2 + \cdots + b_n = \displaystyle \sum_{k=1}^n 2^{k-1}b_1 = c(2^n-1)$

Solve for $c$ .

$2n^2 + 2(c-1)n = c(2^n-1) \implies c = \dfrac{2n^2 - 2n}{2^n - 2n - 1} = \dfrac{2n(n-1)}{2^n - 2n - 1}$

Now, the numerator increases more slowly than the denominator, since

$\dfrac{d}{dn}(2n(n-1)) = 4n - 2$

$\dfrac{d}{dn} (2^n-2n-1) = \ln(2)\cdot2^n - 2$

and for $n\ge5$ ,

$2^n > \dfrac4{\ln(2)} n \implies \ln(2)\cdot2^n - 2 > 4n - 2$

This means we only need to check if the claim is true for any $n\in\{1,2,3,4\}$ .

$n=1$ doesn't work, since that makes $c=0$ .

If $n=2$ , then

$c = \dfrac{4}{2^2 - 4 - 1} = \dfrac4{-1} = -4 < 0$

If $n=3$ , then

$c = \dfrac{12}{2^3 - 6 - 1} = 12$

If $n=4$ , then

$c = \dfrac{24}{2^4 - 8 - 1} = \dfrac{24}7 \not\in\Bbb N$

There is only one value for which the claim is true, $c=12$ .

3 0

2 years ago

In the following problem, check that it is appropriate to use the normal approximation to the binomial. Then use the normal dist

Marrrta [24]

Answer:

a) Bi [P ( X >=15 ) ] ≈ 0.9944

b) Bi [P ( X >=30 ) ] ≈ 0.3182

c) Bi [P ( 25=< X =< 35 ) ] ≈ 0.6623

d) Bi [P ( X >40 ) ] ≈ 0.0046

Step-by-step explanation:

Given:

- Total sample size n = 745

- The probability of success p = 0.037

- The probability of failure q = 0.963

Find:

a. 15 or more will live beyond their 90th birthday

b. 30 or more will live beyond their 90th birthday

c. between 25 and 35 will live beyond their 90th birthday

d. more than 40 will live beyond their 90th birthday

Solution:

- The condition for normal approximation to binomial distribution:

n*p = 745*0.037 = 27.565 > 5

n*q = 745*0.963 = 717.435 > 5

Normal Approximation is valid.

a) P ( X >= 15 ) ?

- Apply continuity correction for normal approximation:

Bi [P ( X >=15 ) ] = N [ P ( X >= 14.5 ) ]

- Then the parameters u mean and σ standard deviation for normal distribution are:

u = n*p = 27.565

σ = sqrt ( n*p*q ) = sqrt ( 745*0.037*0.963 ) = 5.1522

- The random variable has approximated normal distribution as follows:

X~N ( 27.565 , 5.1522^2 )

- Now compute the Z - value for the corrected limit:

N [ P ( X >= 14.5 ) ] = P ( Z >= (14.5 - 27.565) / 5.1522 )

N [ P ( X >= 14.5 ) ] = P ( Z >= -2.5358 )

- Now use the Z-score table to evaluate the probability:

P ( Z >= -2.5358 ) = 0.9944

N [ P ( X >= 14.5 ) ] = P ( Z >= -2.5358 ) = 0.9944

Hence,

Bi [P ( X >=15 ) ] ≈ 0.9944

b) P ( X >= 30 ) ?

- Apply continuity correction for normal approximation:

Bi [P ( X >=30 ) ] = N [ P ( X >= 29.5 ) ]

- Now compute the Z - value for the corrected limit:

N [ P ( X >= 29.5 ) ] = P ( Z >= (29.5 - 27.565) / 5.1522 )

N [ P ( X >= 29.5 ) ] = P ( Z >= 0.37556 )

- Now use the Z-score table to evaluate the probability:

P ( Z >= 0.37556 ) = 0.3182

N [ P ( X >= 29.5 ) ] = P ( Z >= 0.37556 ) = 0.3182

Hence,

Bi [P ( X >=30 ) ] ≈ 0.3182

c) P ( 25=< X =< 35 ) ?

- Apply continuity correction for normal approximation:

Bi [P ( 25=< X =< 35 ) ] = N [ P ( 24.5=< X =< 35.5 ) ]

- Now compute the Z - value for the corrected limit:

N [ P ( 24.5=< X =< 35.5 ) ]= P ( (24.5 - 27.565) / 5.1522 =<Z =< (35.5 - 27.565) / 5.1522 )

N [ P ( 24.5=< X =< 25.5 ) ] = P ( -0.59489 =<Z =< 1.54011 )

- Now use the Z-score table to evaluate the probability:

P ( -0.59489 =<Z =< 1.54011 ) = 0.6623

N [ P ( 24.5=< X =< 35.5 ) ]= P ( -0.59489 =<Z =< 1.54011 ) = 0.6623

Hence,

Bi [P ( 25=< X =< 35 ) ] ≈ 0.6623

d) P ( X > 40 ) ?

- Apply continuity correction for normal approximation:

Bi [P ( X >40 ) ] = N [ P ( X > 41 ) ]

- Now compute the Z - value for the corrected limit:

N [ P ( X > 41 ) ] = P ( Z > (41 - 27.565) / 5.1522 )

N [ P ( X > 41 ) ] = P ( Z > 2.60762 )

- Now use the Z-score table to evaluate the probability:

P ( Z > 2.60762 ) = 0.0046

N [ P ( X > 41 ) ] = P ( Z > 2.60762 ) = 0.0046

Hence,

Bi [P ( X >40 ) ] ≈ 0.0046

4 0

4 years ago