Which type of triangle will always have at least 1-fold reflectional symmetry?

Mathematics

2 answers:

telo118 [61]3 years ago

8 0

Answer: isocleles

Step-by-step explanation:

kodGreya [7K]3 years ago

8 0

Answer:

An Isosceles triangle

Step-by-step explanation:

Isosceles triangle is a type of triangle that will always have exactly 1-fold reflection symmetry.

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Solve the following equation x(x + 19) = -34

xxMikexx [17]

Hi there,

x(x + 19) = -34

I'm going to solve your equation step-by-step.x(x + 19) = −34

Step 1: Simplify both sides of the equation.x2 + 19x = −34

Step 2: Subtract -34 from both sides.x2 + 19x − (−34) = −34 − (−34)

x2 + 19x + 34 = 0

Step 3: Factor left side of equation.(x + 2) (x + 17) = 0

Step 4: Set factors equal to 0.x + 2 = 0 or x + 17 = 0

x = −2 or x = −17

Answer:x = −2 or x = −17

Hope this helps! :)

5 0

3 years ago

Read 2 more answers

g Annual starting salaries in a certain region of the U. S. for college graduates with an engineering major are normally distrib

algol13

Answer:

The probability that the sample mean would be at least $39000 is of 0.8665 = 86.65%.

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 z-score 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 p-value, 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 $s = \frac{\sigma}{\sqrt{n}}$ .