in lebanon, what is the maximum number of the likely subscribers to the chain of 6 digits phone numbers?

? For which value of x is the inequality 45 - 4x < 14 true?

viktelen [127]

Answer:

7.75

Step-by-step explanation:

You group like terms from the inequality above ie by making the x value stands alone and you can only do this by sending 14 to the left side of the inequality and 4x to the right side. This becomes 45-14 < 4x and that gives you 31 < 4x. Then you divide both side by the number attached to the variable 'x' ie 4. So 31/4 < 4x/4 making x > 31/4 which is 7.75

8 0

3 years ago

(1)/(2)(8y-4)=(3)/(4)(16-8y)

Talja [164]

Answer: y=7/5

Hope this helps

7 0

3 years ago

A grocery store’s receipts show that Sunday customer purchases have a skewed distribution with a mean of 27$ and a standard devi

34kurt

Answer:

(a) The probability that the store’s revenues were at least $9,000 is 0.0233.

(b) The revenue of the store on the worst 1% of such days is $7,631.57.

Step-by-step explanation:

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and we take appropriately huge random samples (n ≥ 30) from the population with replacement, then the distribution of the sum of values of X, i.e ∑X, will be approximately normally distributed.

Then, the mean of the distribution of the sum of values of X is given by,

$\mu_{X}=n\mu$

And the standard deviation of the distribution of the sum of values of X is given by,

$\sigma_{X}=\sqrt{n}\sigma$

It is provided that:

$\mu=\$27\\\sigma=\$18\\n=310$

As the sample size is quite large, i.e. n = 310 > 30, the central limit theorem can be applied to approximate the sampling distribution of the store’s revenues for Sundays by a normal distribution.

(a)

Compute the probability that the store’s revenues were at least $9,000 as follows:

$P(S\geq 9000)=P(\frac{S-\mu_{X}}{\sigma_{X}}\geq \frac{9000-(27\times310)}{\sqrt{310}\times 18})\\\\=P(Z\geq 1.99)\\\\=1-P(Z$

Thus, the probability that the store’s revenues were at least $9,000 is 0.0233.

(b)

Let s denote the revenue of the store on the worst 1% of such days.

Then, P (S < s) = 0.01.

The corresponding z-value is, -2.33.

Compute the value of s as follows:

$z=\frac{s-\mu_{X}}{\sigma_{X}}\\\\-2.33=\frac{s-8370}{316.923}\\\\s=8370-(2.33\times 316.923)\\\\s=7631.56941\\\\s\approx \$7,631.57$

Thus, the revenue of the store on the worst 1% of such days is $7,631.57.

5 0

2 years ago

Find each measure.

andreyandreev [35.5K]

Answer:

The BMW E30 is the second generation of BMW 3 Series, which was produced from 1982 to 1994 and replaced the E21 3 Series. The model range included 2-door sedan and convertible body styles,[4][5] as well as being the first 3 Series to be produced in 4-door sedan and wagon/estate body styles. It was powered by four-cylinder petrol, six-cylinder petrol and six-cylinder diesel engines, the latter a first for the 3 Series. The E30 325iX model was the first BMW to have all-wheel drive.

Step-by-step explanation:

4 0

3 years ago

A piece of fabric is 1/3 yard by 3 5/8 yards. Irma solves 1/3⋅3 5/8 to find the area of the fabric in square yards. Enter number

Harlamova29_29 [7]

For this case we have:

The area of the piece of fabric is given by $A = a * b$