3. The following expression is given. 2/3 (3/4x-3/2)

in 2016 the cost of 2 oz of pure gold was $2,640 complete the double number line to show the cost for 1, 3, and 4 oz of gold

Nezavi [6.7K]

Answer:

On the other hand, if rates are very low, gold may potentially benefit as it keeps the opportunity cost of holding gold to a minimum. Of course, gold could also move higher even with high interest rates, and it could move lower even during periods of ultra-low rates. Monetary policy can also affect the gold price.

Step-by-step explanation:

8 0

3 years ago

Y is between X and Z on a line. XZ = 15 and XY = 10. What is the length of YZ?

Mariulka [41]

Answer:

5

Step-by-step explanation:

if xz is 15 and xy is 10 then we can infer that z is 10 and x is 5

3 0

3 years ago

You mow the lawn to earn your allowance. Each time you mow the lawn you earn 12$

BARSIC [14]

Answer:

12×m=d................

5 0

3 years ago

Given the function g(x) = 4(3)x, Section A is from x = 1 to x = 2 and Section B is from x = 3 to x = 4.

LenaWriter [7]

Easy peasy

the average rate of change in section A is the slope from (1,g(1)) to (2,g(2))
the average rate of chagne in section B is the slope from (3,g(3)) to (4,g(4))

A.

section A
g(1)=4(3)^1=12
g(2)=4(3)^2=4(9)=36
slope=(36-12)/(2-1)=24/1=24

section B
g(3)=4(3)^3=4(27)=108
g(4)=4(3)^4=4(81)=324
slope=(324-108)/(4-3)=216/1=216

section A has an average rate of change of 24
section B has an average rate of change of 216

3 0

3 years ago

A report indicated that 37% of adults had received a bogus email intended to steal personal information. Suppose a random sample

fiasKO [112]

Answer:

5.05% probability that no more than 34% had received such an email.

Step-by-step explanation:

We use the binomial approximation to the normal to solve this problem.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be 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.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .