How many 2/5 are in 8?

1 answer:

4 0

Answer : 212
Reason why is because yu have 2/5 you multiply it across 8/1 5x1= 5 and 8x2= 16 then you will have 5/16 you divide 5 into 16 , 5 goes into 16 twice then subtract 16 and 10 then you would have 6 bring down the 0 and you have 60 . 5 goes into 60 12 times then you have your answer 212 .

You might be interested in

01-1-

klio [65]

Answer:

Step-by-step explanation:

50 dollars = ∉34

∉1 = Rs 150

Rs 382800 / 150 Rs per ∉ = 382800/150 ∉

=2552 ∉

50 dollars = ∉34 so 50/34 dollars = ∉

2552 ∉ × 50/34 dollar per ∉ = in the units the ∉ cancel out

∉ × dollar/∉ = dollars

(2552 × 50) / 34 = 3752.94 dollars <------------------

Here is another way of looking at the answer...........

Rs 382800 × 1∉/150 Rs × 50 dollars/34 ∉ = note how the units cancel

= 382800 × (1/150) × (50/34) Rs × ∉/Rs × dollars/∉

= 3,752.94 dollars leaving only dollars in the numerator

7 0

3 years ago

Suppose a batch of metal shafts produced in a manufacturing company have a population standard deviation of 1.3 and a mean diame

lbvjy [14]

Answer:

54.86% probability that the mean diameter of the sample shafts would differ from the population mean by more than 0.1 inches

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

$\mu = 208, \sigma = 1.3, n = 60, s = \frac{1.3}{\sqrt{60}} = 0.1678$

What is the probability that the mean diameter of the sample shafts would differ from the population mean by more than 0.1 inches

Lesser than 208 - 0.1 = 207.9 or greater than 208 + 0.1 = 208.1. Since the normal distribution is symmetric, these probabilities are equal, so we find one of them and multiply by 2.

Lesser than 207.9.

pvalue of Z when X = 207.9. So

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

By the Central Limit Theorem

$Z = \frac{207.9 - 208}{0.1678}$