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2 years ago

PLEAASSSE HELP ASAP I WILL MARK →BRAINLIEST← AND 5 STARS!!!

Mathematics

1 answer:

algol132 years ago

5 0

Answer:Yes she has enough

Step-by-step explanation:

Total ounces=11

1st recipe=6 ounces

2nd recipe=1/4 ounces

1 pound =16 ounces

= 1/4 pounds × 16 ounces/1 pound

= 4 ounces

First recipe + Second recipe

= 6 ounces + 4 ounces

= 10 ounces.

10 ounces of walnuts and Marni has 11

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A convention planner estimates that the size in square feet of a room needed for a large assembly is 20 times the number of atte

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Answer:

4000 Sq feet.

Step-by-step explanation:

Since the room needed is 20 times the number of attendees and 200 attendees are expected to come, it's just 20x200. Looking for keywords such as times really help!

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2 years ago

Earn $134 in 8 hours how much would i get. per hour?

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$16.75
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3 years ago

Read 2 more answers

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

Assume that cost is 2 times more important than reputation and reputation is 3 times more important than compatibility with exis

dusya [7]

Let the weightage of Ease of Use be x
Ease of Use = x

Compatibility is 5 times more than ease of use:
Compatibility = 5x

Reputation is 3 times more important than compatibility:
Reputation = 3(5x)
Reputation = 15x

Cost is 2 times more important than reputation:
Cost = 2(15x)
Cost = 30x

So the weightage are:
Ease of Use : 1
Compatibility : 5
Reputation :15
Cost : 30

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2 years ago

Find the value of each unknown variable in the given figures ?

AveGali [126]

10) a. 130

Step-by-step explanation:

$180 - 50 = 130$

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3 years ago