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3 years ago

Find two consecutive odd numbers whose sum is 44.

Mathematics

1 answer:

SCORPION-xisa [38]3 years ago

3 0

Answer:

21 & 23

<em>Hope that helps!</em>

Step-by-step explanation:

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(7x-11)+(6x-5)+(3x+6)+(4x+7)=

Evgen [1.6K]

the answer is 20x - 3

have a wonderful day! :)

6 0

2 years ago

Simplify the ratio 9/12

Alex777 [14]

3:4 This is the answer

5 0

2 years ago

Read 2 more answers

Jerry had 1 gallon of ice cream he used 3/10 gallon to make chocolate milkshakes and 0.40 gallon to make vanilla milkshakes how

Gre4nikov [31]

Let
x------> the <span>remaining ice cream

we know that
3/10------> 0.30
so

x=1-0.30-0.40-----> x=1-0.70------> x=0.30 gal------> 3/10 gal

the answer is
3/10 gal</span>

6 0

3 years ago

It looks easy i just dont know how to do it can someone help me out

Arada [10]

Answer:

Step-by-step explanation:

16+4x = 10 +14

-16 -16

4x = 8

/4 /4

x=2

8x=8(2)

=16

3 0

2 years ago

For a certain river, suppose the drought length Y is the number of consecutive time intervals in which the water supply remains

AnnZ [28]

Answer:

a) There is a 9% probability that a drought lasts exactly 3 intervals.

There is an 85.5% probability that a drought lasts at most 3 intervals.

b)There is a 14.5% probability that the length of a drought exceeds its mean value by at least one standard deviation

Step-by-step explanation:

The geometric distribution is the number of failures expected before you get a success in a series of Bernoulli trials.

It has the following probability density formula:

$f(x) = (1-p)^{x}p$

In which p is the probability of a success.

The mean of the geometric distribution is given by the following formula:

$\mu = \frac{1-p}{p}$

The standard deviation of the geometric distribution is given by the following formula:

$\sigma = \sqrt{\frac{1-p}{p^{2}}$

In this problem, we have that:

$p = 0.383$

$\mu = \frac{1-p}{p} = \frac{1-0.383}{0.383} = 1.61$

$\sigma = \sqrt{\frac{1-p}{p^{2}}} = \sqrt{\frac{1-0.383}{(0.383)^{2}}} = 2.05$

(a) What is the probability that a drought lasts exactly 3 intervals?

This is $f(3)$

$f(x) = (1-p)^{x}p$

$f(3) = (1-0.383)^{3}*(0.383)$

$f(3) = 0.09$

There is a 9% probability that a drought lasts exactly 3 intervals.

At most 3 intervals?

This is $P = f(0) + f(1) + f(2) + f(3)$

$f(x) = (1-p)^{x}p$

$f(0) = (1-0.383)^{0}*(0.383) = 0.383$

$f(1) = (1-0.383)^{1}*(0.383) = 0.236$

$f(2) = (1-0.383)^{2}*(0.383) = 0.146$

Previously in this exercise, we found that $f(3) = 0.09$

$P = f(0) + f(1) + f(2) + f(3) = 0.383 + 0.236 + 0.146 + 0.09 = 0.855$

There is an 85.5% probability that a drought lasts at most 3 intervals.

(b) What is the probability that the length of a drought exceeds its mean value by at least one standard deviation?

This is $P(X \geq \mu+\sigma) = P(X \geq 1.61 + 2.05) = P(X \geq 3.66) = P(X \geq 4)$ .

We are working with discrete data, so 3.66 is rounded up to 4.

Either a drought lasts at least four months, or it lasts at most thee. In a), we found that the probability that it lasts at most 3 months is 0.855. The sum of these probabilities is decimal 1. So:

$P(X \leq 3) + P(X \geq 4) = 1$

$0.855 + P(X \geq 4) = 1$

$P(X \geq 4) = 0.145$

There is a 14.5% probability that the length of a drought exceeds its mean value by at least one standard deviation

8 0

3 years ago