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3 years ago

Can someone help please

Mathematics

1 answer:

Viefleur [7K]3 years ago

8 0

9514 1404 393

Answer:

A, D

Step-by-step explanation:

The figure <em>looks</em> symmetrical, but their is no given information indicating it actually <em>is</em> symmetrical. Thus we can't make any statements about the segment lengths. Nor is there any indication that any of the arc lengths are identical.

If we rule out statements about segments and arcs, that only leaves statements about angles. As it happens both of those are true: inscribed angles that intercept the same arc are congruent.

The true statements are ...

A: ∠CAD≅∠CBD

D: ∠ACB≅∠ADB

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vekshin1

B) 6, 8, 10

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3 years ago

A digital scale measures weight to the nearest 0.2 pound. Which

AnnyKZ [126]

Answer:

Step-by-step explanation:

This option contains a number that would appear on this scale.

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3 years ago

A person places $82100 in an investment account earning an annual rate of 4.3%, compounded continuously. Using the formula V = P

SVETLANKA909090 [29]

Answer:

$89,472.58

Step-by-step explanation:

P = $82100

r = 4.3% = 4.3/100 = 0.043

T = 2 years

V = P × e^rt

V = 82100 × e^(0.043 × 2)

V = 82100 × e^(0.086)

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The interest after 2 years is $89,472.58

7 0

3 years ago

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exis [7]

9514 1404 393

Answer:

(d) Both sales persons will earn a monthly salary of $5,000 for $25,000 in sales

Step-by-step explanation:

The variable definitions are ...

x = amount of sales

y = monthly earnings

Then the point (x, y) = (25000, 5000) means earnings are $5000 on sales of $25,000.

8 0

3 years ago

(10 points)Assume IQs of adults in a certain country are normally distributed with mean 100 and SD 15. Suppose a president, vice

vesna_86 [32]

Answer:

0.0139 = 1.39% probability that the president will have an IQ of at least 107.5 and that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130.

Step-by-step explanation:

To solve this question, we need to use the binomial and the normal probability distributions.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

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.

Probability the president will have an IQ of at least 107.5

IQs of adults in a certain country are normally distributed with mean 100 and SD 15, which means that $\mu = 100, \sigma = 15$

This probability is 1 subtracted by the p-value of Z when X = 107.5. So

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

$Z = \frac{107.5 - 100}{15}$

$Z = 0.5$

$Z = 0.5$ has a p-value of 0.6915.

1 - 0.6915 = 0.3085

0.3085 probability that the president will have an IQ of at least 107.5.

Probability that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130.

First, we find the probability of a single person having an IQ of at least 130, which is 1 subtracted by the p-value of Z when X = 130. So

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

$Z = \frac{130 - 100}{15}$

$Z = 2$

$Z = 2$ has a p-value of 0.9772.

1 - 0.9772 = 0.0228.

Now, we find the probability of at least one person, from a set of 2, having an IQ of at least 130, which is found using the binomial distribution, with p = 0.0228 and n = 2, and we want:

$P(X \geq 1) = 1 - P(X = 0)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{2,0}.(0.9772)^{2}.(0.0228)^{0} = 0.9549$

$P(X \geq 1) = 1 - P(X = 0) = 0.0451$

0.0451 probability that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130.

What is the probability that the president will have an IQ of at least 107.5 and that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130?

0.3085 probability that the president will have an IQ of at least 107.5.

0.0451 probability that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130.

Independent events, so we multiply the probabilities.

0.3082*0.0451 = 0.0139

8 0

3 years ago