What is 219÷3 help please

The distribution of the number of transactions performed at a bank each day is approximately normal with mean 478 transactions a

jolli1 [7]

Answer:

e. 0.977

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 478, \sigma = 64$

Which of the following is closest to the proportion of daily transactions greater than 350?

This is 1 subtracted by the pvalue of Z when X = 350. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{350 - 478}{64}$

$Z = -2$

$Z = -2$ has a pvalue of 0.023

1 - 0.023 = 0.977

So the correct answer is:

e. 0.977

6 0

3 years ago

Find a solution to the inequality x > 4. A. 8 B. 3 C. 4 D. 0

ella [17]

Answer:

8

Step-by-step explanation:

Isn't it simple because all other numbers are less than 4?

3 0

2 years ago

Joe decides to take his six year old son to the planetarium. The price for the child's ticket is 5.75 dollars less than the pric

zaharov [31]

It would be x + 5.75 because the child ticket is 5.75 less than the adult ticket so to solve how to get the adult ticket you do the opposite

6 0

3 years ago

Read 2 more answers

A swimming pool whose volume is 10 comma 000 gal contains water that is 0.03% chlorine. Starting at tequals0, city water conta

Katarina [22]

Answer:

C(60) = 2.7*10⁻⁴

t = 1870.72 s

Step-by-step explanation:

Let x(t) be the amount of chlorine in the pool at time t. Then the concentration of chlorine is

C(t) = 3*10⁻⁴*x(t).

The input rate is 6*(0.001/100) = 6*10⁻⁵.

The output rate is 6*C(t) = 6*(3*10⁻⁴*x(t)) = 18*10⁻⁴*x(t)

The initial condition is x(0) = C(0)*10⁴/3 = (0.03/100)*10⁴/3 = 1.

The problem is to find C(60) in percents and to find t such that 3*10⁻⁴*x(t) = 0.002/100.

Remember, 1 h = 60 minutes. The initial value problem is

dx/dt= 6*10⁻⁵ - 18*10⁻⁴x = - 6* 10⁻⁴*(3x - 10⁻¹) x(0) = 1.

The equation is separable. It can be rewritten as dx/(3x - 10⁻¹) = -6*10⁻⁴dt.

The integration of both sides gives us

Ln |3x - 0.1| / 3 = -6*10⁻⁴*t + C or |3x - 0.1| = e∧(3C)*e∧(-18*10⁻⁴t).

Therefore, 3x - 0.1 = C₁*e∧(-18*10⁻⁴t).

Plug in the initial condition t = 0, x = 1 to obtain C₁ = 2.9.

Thus the solution to the IVP is

x(t) = (1/3)(2.9*e∧(-18*10⁻⁴t)+0.1)

then

C(t) = 3*10⁻⁴*(1/3)(2.9*e∧(-18*10⁻⁴t)+0.1) = 10⁻⁴*(2.9*e∧(-18*10⁻⁴t)+0.1)

If t = 60

We have

C(60) = 10⁻⁴*(2.9*e∧(-18*10⁻⁴*60)+0.1) = 2.7*10⁻⁴

Now, we obtain t such that 3*10⁻⁴*x(t) = 2*10⁻⁵

3*10⁻⁴*(1/3)(2.9*e∧(-18*10⁻⁴t)+0.1) = 2*10⁻⁵

t = 1870.72 s

7 0

3 years ago

−2(2x+3)≤26 - 2 2 x + 3 ≤ 26 ?

SashulF [63]

$\huge \boxed{\mathbb{QUESTION} \downarrow}$

$- 2 ( 2 x + 3 ) \leq 26.$

$\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}$

$- 2 ( 2 x + 3 ) \leq 26$

Divide both sides by -2. We know that since 2 is negative, it'll be < (less than) 0. So, the direction of inequality will change.

$2x+3\geq \frac{26}{-2} \\$

Divide 26 by -2..we'll get the answer as -13.

$2x+3\geq -13$

Now, subtract 3 from both the sides.

$2x\geq -13-3$

Subtracting -13 by 3..we'll get the answer as -16.

$2x\geq -16$

Divide both sides by 2. We know that since 2 is positive, the direction of inequality will remain the same.

$x\geq \frac{-16}{2} \\$

Divide -16 by 2..we'll get the answer as -8.

$\huge \boxed{ \boxed{ \bf \: x\geq -8 }}$

The correct answer is ⇨ x greater than or equal to -8.

3 0

2 years ago