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2 years ago

For each set of three measures determine if they can be angle measures of a triangle

Mathematics

1 answer:

Orlov [11]2 years ago

8 0

Only the second set of measures qualifies as angle measures of a triangle. Angles 25°, 130°, 25°

Step-by-step explanation:

Step 1: To find whether angles qualify as angle measures of a triangles, calculate their total sum and verify whether they add up to 180°

Set 1 - 41° + 112° + 52° = 205°

Set 2 - 25° + 130° + 25° = 180°

Set 3 - 30° + 40° + 90° = 160°

Set 4 - 132° + 141° + 31° = 304°

Therefore, only set 2 qualifies.

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please help solve this problem the - is a negative sign. (2b+-7)+5(b+-3)

enyata [817]

Answer:

Step-by-step explanation:

(2b+-7)+5(b+-3)

Multiply 5 throughout the second parentheses

(2b-7)+(5b-15)

Add 2b and 5b

7b-7-15

Subtract -7 and -15

7b-22

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3 years ago

In a 1995 Corporation for Public Broadcasting poll of TV viewership, one question was, "A recent study by a psychology professor

sweet [91]

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2 years ago

Guys, Can you please help me with these questions

MakcuM [25]

Answer:

a) $w^{13} x^{5} y^{6}$

b) $\frac{x}{3y^{6} }$

Step-by-step explanation:

a) $(w^{2} xy^{3} )^{2}(w^{3}x )^{3}$

1. Distribute the second power (2) outside the first pair of parenthesis:

$(w^{2(2)} x^{2} y^{3(2)} )$

= $w^{4} x^{2} y^{6} (w^{3}x )^{3}$

2. Distribute the third power (3) outside the second pair of parenthesis:

$(w^{3(3)} x^{3} )$

= $w^{4} x^{2} y^{6} w^{9} x^{3}$

3. Combine like terms:

$w^{13} x^{5} y^{6}$

--------------------------------------------

b) $\frac{2x^{2} y^{5} }{6xy^{11} }$

1. Factor the number 6 (= 2 · 3):

$\frac{2x^{2} y^{5} }{2(3)xy^{11} }$

2. Cancel the common factor (2):

$\frac{x^{2} y^{5} }{3xy^{11} }$

3. Cancel out $xy^{5}$ in the numerator an denominator:

$\frac{x}{3y^{6} }$

hope this helps!

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2 years ago

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3 years ago

. Exam scores for a large introductory statistics class follow an approximate normal distribution with an average score of 56 an

Andru [333]

Answer:

0.1% probability that the average score of a random sample of 20 students exceeds 59.5.

Step-by-step explanation:

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 $\frac{\sigma}{\sqrt{n}}$

Normal probability distribution

Problems of normally distributed samples 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.

In this problem, we have that:

$\mu = 56, \sigma = 5, n = 20, s = \frac{5}{\sqrt{20}} = 1.12$