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2 years ago

Find the length of the segment indicated. Round your answer to the nearest tenth if necessary.

Mathematics

1 answer:

Elena-2011 [213]2 years ago

4 0

Given:

The length of the segment of the chord DB is 8.2 units.

The length of the segment AB is 6.9 units.

The length of the radius AC be x units.

We need to determine the value of x.

Length of BC:

Since, we know the property that, "if a radius is perpendicular to the chord, then it bisects the chord".

Thus, applying the above property, we have;

DB ≅ BC

8.2 = BC

Thus, the length of BC is 8.2 units.

Value of x:

Since, ∠B makes 90°, let us apply the Pythagorean theorem to determine the value of x.

Thus, we have;

$AC^2=AB^2+BC^2$

Substituting the values, we have;

$x^2=6.9^2+8.2^2$

$x^2=47.61+67.24$

$x^{2} =114.85$

$x=10.7$

Thus, the value of x is 10.7 units.

Hence, Option A is the correct answer.

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Triangle E F G is shown. Which statements regarding Triangle E F G are true? Select three options. E F + F G greater-than E G E

Leokris [45]

Answer:

options 1 ( E F + F G greater-than E G ), 2 (E G + F G greater-than E F ) and 5 ( E G + E F less-than F G ) are true.

Step-by-step explanation:

Given triangle EFG.

In each tirangle with sides lengths a, b and c, we always have

$a+b>c\\ \\a+c>b\\ \\b+c>a$

In triangle EFG:

$EF+FG>EG;$
$EG+GF>EF;$
$FE+EG>FG.$

So, options 1 ( E F + F G greater-than E G ), 2 (E G + F G greater-than E F ) and 5 ( E G + E F less-than F G ) are true.

Options 3 and 4 are false.

4 0

2 years ago

Read 2 more answers

A university surveyed recent graduates of the English department for their starting salaries. Four hundred graduates returned th

NeX [460]

Answer:

Step-by-step explanation:

We want to determine a 95% confidence interval for the mean salary of all graduates from the English department.

Number of sample, n = 400

Mean, u = $25,000

Standard deviation, s = $2,500

For a confidence level of 95%, the corresponding z value is 1.96. This is determined from the normal distribution table.

We will apply the formula

Confidence interval

= mean ± z × standard deviation/√n

It becomes

25000 ± 1.96 × 2500/√400

= 25000 ± 1.96 × 125

= 25000 ± 245

The lower end of the confidence interval is 25000 - 245 =24755

The upper end of the confidence interval is 25000 + 245 = 25245

Therefore, with 95% confidence interval, the mean salary of all graduates from the English department is between $24755 and $25245

3 0

2 years ago

A particular telephone number is used to receive both voice calls and fax messages. Suppose that 25% of the incoming calls invol

bagirrra123 [75]

Answer:

a) 0.214 = 21.4% probability that at most 4 of the calls involve a fax message

b) 0.118 = 11.8% probability that exactly 4 of the calls involve a fax message

c) 0.904 = 90.4% probability that at least 4 of the calls involve a fax message

d) 0.786 = 78.6% probability that more than 4 of the calls involve a fax message

Step-by-step explanation:

For each call, there are only two possible outcomes. Either it involves a fax message, or it does not. The probability of a call involving a fax message is independent of other calls. So we use the binomial probability distribution to solve this question.

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.

25% of the incoming calls involve fax messages

This means that $p = 0.25$

25 incoming calls.

This means that $n = 25$

a. What is the probability that at most 4 of the calls involve a fax message?

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$ .

In which

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

$P(X = 0) = C_{25,0}.(0.25)^{0}.(0.75)^{25} = 0.001$

$P(X = 1) = C_{25,1}.(0.25)^{1}.(0.75)^{24} = 0.006$

$P(X = 2) = C_{25,2}.(0.25)^{2}.(0.75)^{23} = 0.025$

$P(X = 3) = C_{25,3}.(0.25)^{3}.(0.75)^{22} = 0.064$

$P(X = 4) = C_{25,4}.(0.25)^{4}.(0.75)^{21} = 0.118$

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.001 + 0.006 + 0.025 + 0.064 + 0.118 = 0.214$

0.214 = 21.4% probability that at most 4 of the calls involve a fax message

b. What is the probability that exactly 4 of the calls involve a fax message?

$P(X = 4) = C_{25,4}.(0.25)^{4}.(0.75)^{21} = 0.118$

0.118 = 11.8% probability that exactly 4 of the calls involve a fax message.

c. What is the probability that at least 4 of the calls involve a fax message?

Either less than 4 calls involve fax messages, or at least 4 do. The sum of the probabilities of these events is 1. So

$P(X < 4) + P(X \geq 4) = 1$

We want $P(X \geq 4)$ . Then

$P(X \geq 4) = 1 - P(X < 4)$

In which

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

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

$P(X = 0) = C_{25,0}.(0.25)^{0}.(0.75)^{25} = 0.001$

$P(X = 1) = C_{25,1}.(0.25)^{1}.(0.75)^{24} = 0.006$

$P(X = 2) = C_{25,2}.(0.25)^{2}.(0.75)^{23} = 0.025$

$P(X = 3) = C_{25,3}.(0.25)^{3}.(0.75)^{22} = 0.064$

$P(X$

$P(X \geq 4) = 1 - P(X < 4) = 1 - 0.096 = 0.904$

0.904 = 90.4% probability that at least 4 of the calls involve a fax message.

d. What is the probability that more than 4 of the calls involve a fax message?

Very similar to c.

$P(X \leq 4) + P(X > 4) = 1$

From a), $P(X \leq 4) = 0.214)$

Then

$P(X > 4) = 1 - 0.214 = 0.786$

0.786 = 78.6% probability that more than 4 of the calls involve a fax message

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2 years ago

A lotion is made from an oil blend costing $1.50 per ounce and glycerin costing $1.00 per ounce. Four ounces of lotion costs $5.

Svet_ta [14]

Answer:

A (4-g)1.5

Step-by-step explanation:

hope this helps if you need a explanation just ask in the comments and I will.

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2 years ago

a school baseball team raised $810 for new uniforms. each player on the team sold one book of tickets. There were 10 tickets in

antiseptic1488 [7]

27
Multiply 10 and 3 because thats how much each book is worth.
Then divide 810 by 30
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2 years ago