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3 years ago

Solve: a. 5y – 3 = – 18 b. -3x – 9 = 0 c. 4 + 3(z - 8) = -23 d. 1 – 2(y – 4) = 5

Mathematics

1 answer:

oksian1 [2.3K]3 years ago

3 0

Answer:

a. y = -3

b. x = -3

c. z = - 1

d. y = -7

Step-by-step explanation:

a. 5y - 3 = - 18

5y = -18 +3

5y = -15

y = -15/5

y = - 3

b. -3x - 9 = 0

-3x = 9

x = 9/-3

x = -3

c. 4 + 3(z-8)= -23

4 + 3z-24 = -23

3z - 20 = -23

3z = -23 + 20

3z = -3

z = -3/3

z = - 1

d. 1 - 2(y-4) = 5

1 - 2y + 8 = 5

9 - 2y = 5

-2y = 14

y = 14/-2

y = -7

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In a population of similar households, suppose the weekly supermarket expense for a typical household is normally distributed wi

Rina8888 [55]

Answer:

P(Y ≥ 15) = 0.763

Step-by-step explanation:

Given that:

Mean =135

standard deviation = 12

sample size n = 50

sample mean $\overline x$ = 140

Suppose X is the random variable that follows a normal distribution which represents the weekly supermarket expenses

Then,

$X \sim N ( \mu \sigma)$

The probability that X is greater than 140 is :

P(X>140) = 1 - P(X ≤ 140)

$P(X>140) = 1 - P( \dfrac{X-\mu}{\sigma} \leq \dfrac{140-135}{12})$

$P(X>140) = 1 - P( \dfrac{X-\mu}{\sigma} \leq \dfrac{5}{12})$

$P(X>140) = 1 - P( Z\leq0.42)$

From z tables,

$P(X>140) = 1 - 0.6628$

$P(X>140) = 0.3372$

Similarly, let consider Y to be the variable that follows a binomial distribution of the no of household whose expense is greater than $140

Then;

$Y \sim Binomial (np)$

$Y \sim Binomial (50,0.3372)$

∴

P(Y ≥ 15) = 1- P(Y< 15)

P(Y ≥ 15) = 1 - ( P(Y=0) + P(Y=1) + P(Y=2) + ... + P(Y=14) )

$P(Y \geq 15) = 1 - \begin {pmatrix} ^{50}_0 \end {pmatrix} (0.3372)^0 (1-0,3372)^{50} + \begin {pmatrix} ^{50}_1 \end {pmatrix} (0.3372)^1 (1-0,3372)^{49} + \begin {pmatrix} ^{50}_2 \end {pmatrix} (0.3372)^2 (1-0,3372)^{48} +... + \begin {pmatrix} ^{50}_{50{ \end {pmatrix} (0.3372)^{50} (1-0,3372)^{0}$