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2 years ago

If m∠ABD=39 and m∠DBC=22 what is m∠ABC PLEASE HELP!!!!

Mathematics

1 answer:

Mandarinka [93]2 years ago

5 0

Answer:

∠ABC = 61°

Step-by-step explanation:

Given:

∠ABD = 39°

∠DBC = 22°

Find:

∠ABC = ?

Computation:

We now that,

∠ABC = ∠ABD + ∠DBC

∠ABC = 39° + 22°

∠ABC = 61°

Therefore, ∠ABC is 61°

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Find the perimeter of the C shape?

Ierofanga [76]

68 meters is the correct answer

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3 years ago

F(x)=|x+8| evaluate f(-11)

stealth61 [152]

19:)
I hope that helped!

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3 years ago

A fair coin is to be tossed 20 times. Find the probability that 10 of the tosses will fall heads and 10 will fall tails, (a) usi

lbvjy [14]

Using the distributions, it is found that there is a:

a) 0.1762 = 17.62% probability that 10 of the tosses will fall heads and 10 will fall tails.

b) 0% probability that 10 of the tosses will fall heads and 10 will fall tails.

c) 0.1742 = 17.42% probability that 10 of the tosses will fall heads and 10 will fall tails.

Item a:

Binomial probability distribution

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

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

The parameters are:

x is the number of successes.
n is the number of trials.
p is the probability of a success on a single trial.

In this problem:

20 tosses, hence $n = 20$ .
Fair coin, hence $p = 0.5$ .

The probability is P(X = 10), thus:

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

$P(X = 10) = C_{20,10}.(0.5)^{10}.(0.5)^{10} = 0.1762$

0.1762 = 17.62% probability that 10 of the tosses will fall heads and 10 will fall tails.

Item b:

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

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

It 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, which is the percentile of X.
The binomial distribution is the probability of x successes on n trials, with p probability of a success on each trial. It can be approximated to the normal distribution with $\mu = np, \sigma = \sqrt{np(1-p)}$ .

The probability of an exact value is 0, hence 0% probability that 10 of the tosses will fall heads and 10 will fall tails.

Item c:

For the approximation, the mean and the standard deviation are:

$\mu = np = 20(0.5) = 10$

$\sigma = \sqrt{np(1 - p)} = \sqrt{20(0.5)(0.5)} = \sqrt{5}$

Using continuity correction, this probability is $P(10 - 0.5 \leq X \leq 10 + 0.5) = P(9.5 \leq X \leq 10.5)$ , which is the p-value of Z when X = 10.5 subtracted by the p-value of Z when X = 9.5.

X = 10.5:

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

$Z = \frac{10.5 - 10}{\sqrt{5}}$

$Z = 0.22$

$Z = 0.22$ has a p-value of 0.5871.

X = 9.5:

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

$Z = \frac{9.5 - 10}{\sqrt{5}}$

$Z = -0.22$

$Z = -0.22$ has a p-value of 0.4129.

0.5871 - 0.4129 = 0.1742.

0.1742 = 17.42% probability that 10 of the tosses will fall heads and 10 will fall tails.

A similar problem is given at brainly.com/question/24261244

6 0

2 years ago

Please answer me without spamming

brilliants [131]

THE ANSWER :- h = 93.88 cm

Step by step:-

6 0

3 years ago

The marked price of a radio set is 450000. A discount of 5% of the marked price is allowed. What is the selling price of the rad

dybincka [34]

Answer:

$427,500 dollars

Step-by-step explanation:

Given the following question:
Initial price = 450,000
5% discount

In order to find the answer, we simply find 5% of 450,000.

$\frac{5\times450000}{100} =5\times450000=2250000\div100=22500$
$450000-22500=427500$
$r=427500$

The radio (after a 5% discount) costs "$427,500 dollars."

Hope this helps.

6 0

2 years ago