Find the measure of the indicated angel

Triangle ABC is similar to triangle DEF. The length of AC is 10cm. The length of BC is 16 cm. The length of DF is 8cm. What is t

Zolol [24]

Answer: The length of EF=15 cm.

Step-by-step explanation:

8 0

3 years ago

You guys are worth it!

Alik [6]

Answer:

i know im amazing

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Multiply the polynomial (12x^4 - 4x^3 + 5) by (-4x^2) *

murzikaleks [220]

Answer:

 $- 48 {x}^{6} + 16 {x}^{5} - 20 {x}^{2}$ 

Option D is the correct option.

Solution,

 $- 4 {x}^{2} (12 {x}^{4} - 4 {x}^{3} + 5) \\ = - 4 {x}^{2} \times 12 {x}^{4} + 4 {x}^{2} \times 4 {x}^{3} - 4 {x}^{2} \times 5 \\ = - 48 {x}^{6} + 16 {x}^{5} - 20 {x}^{2}$ 

hope this helps...

Good luck on your assignment...

7 0

3 years ago

Reduce the Fractions 15/9to its Lowest Terms

KatRina [158]

$\large\bf{\red{ \implies}} \tt \: \cancel\frac{15}{9} \\$

$\large\bf{\red{ \implies}} \tt \: \frac{5}{3} \\$

Decimal Form - 1.6667

6 0

3 years ago

CNNBC recently reported that the mean annual cost of auto insurance is 965 dollars. Assume the standard deviation is 113 dollars

velikii [3]

Answer:

P(939.6 < X < 972.5) = 0.6469

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