2. How many cups of flour are needed to make 6 cupcakes?

In the figure below what is the value of x?

vazorg [7]

Answer:

The value of x is 106

Step-by-step explanation:

Let us solve the question

In the given figure

∵ The triangle in the given figure has two sides equal

→ That means the triangle is an isosceles triangle

∴ The triangle is isosceles

∵ In the isosceles triangle, the base angles are equal in measures

∵ The measure of one base angle is 37°

∴ The measure of other base angle is 37°

→ The sum of the measures of the interior angles of a Δ is 180°

∵ 37° + 37° + x = 180°

→ Add the like terms

∴ 74° + x = 180°

→ Subtract 74 from both sides

∵ 74 - 74 + x = 180 - 74

∴ x = 106

∴ The value of x is 106

7 0

3 years ago

The mean per capita income is 16,127 dollars per annum with a variance of 682,276. What is the probability that the sample mean

MakcuM [25]

Answer:

0.60% probability that the sample mean would differ from the true mean by more than 104 dollars if a sample of 476 persons is randomly selected

Step-by-step explanation:

To solve this question, we have 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 random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean \mu and standard deviation, which is also called standard error $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

The standard deviation is the square root of the variance. So

$\mu = 16127, \sigma = \sqrt{682276} = 826, n = 476, s = \frac{826}{\sqrt{476}} = 37.86$

What is the probability that the sample mean would differ from the true mean by more than 104 dollars if a sample of 476 persons is randomly selected

Either it differs by 104 or less dollars, or it differs by more than 104 dollars. The sum of the probabilities of these events is 100. I am going to find the probability that it differs by 104 or less dollars first.

Probability that it differs by 104 or less dollars first.

pvalue of Z when X = 16127 + 104 = 16231 subtracted by the pvalue of Z when X = 16127 - 104 = 16023. So

X = 16231

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

By the Central Limit Theorem

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

$Z = \frac{16231 - 16127}{37.86}$

$Z = 2.75$

$Z = 2.75$ has a pvalue of 0.9970

X = 16023

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

$Z = \frac{16023 - 16127}{37.86}$

$Z = -2.75$

$Z = -2.75$ has a pvalue of 0.0030

0.9970 - 0.0030 = 0.9940

99.40% probability that it differs by 104 or less.

What is the probability that the sample mean would differ from the true mean by more than 104 dollars if a sample of 476 persons is randomly selected

p + 99.40 = 100

p = 0.60

0.60% probability that the sample mean would differ from the true mean by more than 104 dollars if a sample of 476 persons is randomly selected

7 0

3 years ago

Its for science but its got math involved its got to do with speed distance and time

KIM [24]

distance:1, 1, 1.3, 1.5, 2, 2.3, 2.5, 3 ,3.2, 4

time. :1, 2, 3 , 4 , 5 , 6 , 7 , 8 ,. 9 ,10

distance:1.5, 2, 3.2, 4.2, 5, 6, 7, 8

time. : 1 , 2, 3 , 4. , 5, 6, 7, 8

<h3>at 5 seconds which travelled the greatest distance : snail 2</h3>