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Leno4ka [110]
4 years ago
11

A triangle’s first angle measure is 72∘.

Mathematics
1 answer:
yan [13]4 years ago
8 0

An Acute triangle because none of the angels are above 90

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X^2-x-12 factor completely
steposvetlana [31]
Use the sum-product pattern

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x2−x−12

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x2+3x−4x−12
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Common factor from the two pairs

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x2+3x−4x−12

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x(x+3)−4(x+3)
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Rewrite in factored form

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(x−4)(x+3)
5 0
3 years ago
???!!!!????????????
lana [24]
2x=720-(120+120+120+100+10)
2x=720-470
2x=250
simplify by dividing each side by 20
answer: x=125
5 0
3 years ago
Read 2 more answers
2/5 of 2/3
Vilka [71]

Answer:

4/15 is your answer my friend

Step-by-step explanation:

5 0
3 years ago
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Three baskets contain three consecutive odd numbers of apples. There are 105 apples in total. How many apples are in each basket
galben [10]
There are 35 apples in each basket. 105/3 is 35. i got 3 by the three baskets, and 35 is a odd number which also adds up to 105.
8 0
3 years ago
A process is producing a particular part where the thickness of the part is following a normal distribution with a µ = 50 mm and
Hitman42 [59]

Answer:

0.13% probability that this selected sample has an average thickness greater than 53

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 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 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}}.

In this problem, we have that:

\mu = 50, \sigma = 5, n = 25, s = \frac{5}{\sqrt{25}} = 1

What is the probability that this selected sample has an average thickness greater than 53?

This is 1 subtracted by the pvalue of Z when X = 53. So

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

By the Central Limit Theorem

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

Z = \frac{53 - 50}{1}

Z = 3

Z = 3 has a pvalue of 0.9987

1 - 0.9987 = 0.0013

0.13% probability that this selected sample has an average thickness greater than 53

5 0
3 years ago
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