What is the general form of the equation for the given circle centered at O(0, 0)?

An airplane is flying at an altitude of 33,000 feet. It needs to reduce its altitude by 6% to avoid some turbulence. After it pa

Tema [17]

Step-by-step explanation:

100% = 33,000 ft

1% = 100%/100 = 33000/100 = 330 ft

6% = 1% × 6 = 330 × 6 = 1980 ft

that means the plane has to go down 1980 ft to

33,000 - 1980 = 31,020 ft

and then it goes up by 5000 ft :

31,020 + 5000 = 36,020 ft

that is the final altitude.

5 0

2 years ago

Read 2 more answers

What is the least common multiple of 4 and 8

Ray Of Light [21]

Answer: 8

Step-by-step explanation: To find the least common multiple or (lcm) of 4 and 8, let's begin by listing the first few multiples of each number.

Multiples of 4

1 × 4 ≈ 4

2 × 4 ≈ 8

3 × 4 ≈ 12

4 × 4 ≈ 16

Notice that we skipped 0 × 6 in our list of multiples and this is because 0 × 4 is equal to 0 and our least common multiple can't be 0.

Now, let's list the multiples of 8. When listing the multiples of 8, it's a good idea to keep an eye on the list of multiples for 4 so that we will notice when we find a least common multiple.

Multiples for 8

1 × 8 = 8 ← is a multiple of 4

Notice that we can stop here because all other multiples that we find will be greater than 8. Therefore, the least common multiple or (lcm) of 4 and 8 is 8.

7 0

3 years ago

Please helllppppppppp

kondaur [170]

Answer:

x=3cm

Step-by-step explanation:

This is a basic proportion question.

2/18 = x/27

4 0

3 years ago

Please help! For 12 points!

sergij07 [2.7K]

I think it would be the first one and the last one

7 0

3 years ago

In the following problem, check that it is appropriate to use the normal approximation to the binomial. Then use the normal dist

frosja888 [35]

Answer:

a) 0.9920 = 99.20% probability that 15 or more will live beyond their 90th birthday

b) 0.2946 = 29.46% probability that 30 or more will live beyond their 90th birthday

c) 0.6273 = 62.73% probability that between 25 and 35 will live beyond their 90th birthday

d) 0.0034 = 0.34% probability that more than 40 will live beyond their 90th birthday

Step-by-step explanation:

We solve this question using the normal approximation to the binomial distribution.

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 distributions 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:

Sample of 723, 3.7% will live past their 90th birthday.

This means that $n = 723, p = 0.037$ .

So for the approximation, we will have:

$\mu = E(X) = np = 723*0.037 = 26.751$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{723*0.037*0.963} = 5.08$

(a) 15 or more will live beyond their 90th birthday

This is, using continuity correction, $P(X \geq 15 - 0.5) = P(X \geq 14.5)$ , which is 1 subtracted by the pvalue of Z when X = 14.5. So

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

$Z = \frac{14.5 - 26.751}{5.08}$

$Z = -2.41$

$Z = -2.41$ has a pvalue of 0.0080

1 - 0.0080 = 0.9920

0.9920 = 99.20% probability that 15 or more will live beyond their 90th birthday

(b) 30 or more will live beyond their 90th birthday

This is, using continuity correction, $P(X \geq 30 - 0.5) = P(X \geq 29.5)$ , which is 1 subtracted by the pvalue of Z when X = 29.5. So

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

$Z = \frac{29.5 - 26.751}{5.08}$

$Z = 0.54$

$Z = 0.54$ has a pvalue of 0.7054

1 - 0.7054 = 0.2946

0.2946 = 29.46% probability that 30 or more will live beyond their 90th birthday

(c) between 25 and 35 will live beyond their 90th birthday

This is, using continuity correction, $P(25 - 0.5 \leq X \leq 35 + 0.5) = P(X 24.5 \leq X \leq 35.5)$ , which is the pvalue of Z when X = 35.5 subtracted by the pvalue of Z when X = 24.5. So