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Harrizon [31]
3 years ago
5

What type of angle is ∠CEB ? acute right obtuse straight

Mathematics
2 answers:
anygoal [31]3 years ago
7 0
BED and CEB are supplementary angles, meaning their sum is 180°. You already know BED is 77°, so you subtract 77° from 180° to find CEB.

180° - 77° = 103°

CEB = 103°

Because it is greater than 90°, it is obtuse.
Rudik [331]3 years ago
4 0
∠CEB and ∠AED are vertical angles, which means they must be equal to each other. Make them = x in an equation. The angles form a quadrilateral, so the sum of the interior angles must equal 360°.

77 + 77 + x + x = 360
Combine like terms.

154 + 2x = 360
Subtract 154 from both sides.

2x = 206
Divide both sides by 2.

x = 103

∠CEB is greater than 90°, so it must be an obtuse angle.
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Answer:

x = 2 or 1/4

Step-by-step explanation:

-13/4 -x= 1/2x -1

Collect like terms

-13/4+1=1/2x+x

Using LCM

(-13+4)/4=(1+2x²)/2x

9/4=(1+2x²)/2x

Cross multiply

9(2x)=4(1+2x²)

18x=4+8x²

Turn into quadratic and solve

8x²-18x+4

Using formulae method

-b±(√b²-4ac)/2a

Where a=8, b= -18 and c=4

(-(-18)±(√(-18)²-4(8)(4))/2(8)

(18±(√324-128))/16

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(18±14)/16

(18+14)/16 or (18-14)/16

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larisa86 [58]
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What is the slope that goes through (6,-7),(-5,-2)
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An article written for a magazine claims that 78% of the magazine's subscribers report eating healthily the previous day. Suppos
Schach [20]

Answer:

89.44% probability that less than 80% of the sample would report eating healthily the previous day

Step-by-step explanation:

We use the binomial approximation to the normal to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

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.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

p = 0.78, n = 675

So

\mu = E(X) = np = 675*0.78 = 526.5

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{675*0.78*0.22} = 10.76

What is the approximate probability that less than 80% of the sample would report eating healthily the previous day?

This is the pvalue of Z when X = 0.8*675 = 540. So

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

Z = \frac{540 - 526.5}{10.76}

Z = 1.25

Z = 1.25 has a pvalue of 0.8944

89.44% probability that less than 80% of the sample would report eating healthily the previous day

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What is the surface area of a sphere with a radius of 19 units?
almond37 [142]

Hope this helps. Please mark brainliest if it did

Answer: 4536.46

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