There are___puppies that weigh less than 102 g

musickatia [10]

Answer:

x = 46

Step-by-step explanation:

Given 2 secants from an external point to the circle, then

The product of the external part and the whole of one secant is equal to the product of the external part and the whole of the other secant, that is

2(2 + x) = 6(6 + 10)

2(2 + x) = 6 × 16 = 96 ( divide both sides by 2 )

2 + x = 48 ( subtract 2 from both sides )

x = 46

5 0

2 years ago

Which expression is equivalent to 7/12?

Luden [163]

if i understand correctly what you mean by this question the answer could be

14/24

28/48

you can just keep multiplying the top and the bottom by the same number to get an equivalent fraction

im sorry if i understood the question wrong, i hope this helped a little bit

4 0

2 years ago

Read 2 more answers

Concerns about the climate change and CO2 reduction have initiated the commercial production of blends of biodiesel (e.g. from r

natta225 [31]

Answer:

a) 99% of the sample means will fall between 0.933 and 0.941.

b) By the Central Limit Theorem, approximately normal, with mean 0.937 and standard deviation 0.0015.

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 normal distributions 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.

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 $s = \frac{\sigma}{\sqrt{n}}$ .