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3 years ago

The odds against a particular event happening are 9 : 7. What is the probability that the event will happen?

Mathematics

1 answer:

Vsevolod [243]3 years ago

7 0

Answer:

7/16

Step-by-step explanation:

Odds against = number of failures / number of successes

So number of failures = 9 and the number of successes = 7.

Probability = number of successes / number of total events

P = 7 / (7 + 9)

P = 7/16

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Jaxon has $65 in his savings account. He deposits $15 every week. His father also deposits $25 into the account every time Jaxon

Grace [21]

Part A:
A coefficient can be either '15' or '25'.
A variable can be either 'w' or 'm'.
A constant is 65.

Part B:
Simply substitute, or plug-in, the numbers and solve.
***Step 1:
65+15w+25m --> 65+15(20)+25(3)
You do this because you are substituting the 'w' for the number
of weeks that Jaxon saved up for, which is 20, and the 'm' for the number
of times that Jaxon mowed the lawn, which is 3.
***Step 2:
65+15(20)+25(3) --> 65+300+75
Begin to solve, using PEMDAS, or whichever acronym you learned.
Remember, if you are using PEMDAS, recall that the order is Parenthesis,
Exponents, Multiplication/Division (whichever comes first), and
Addition/Subtraction (whichever comes first). Here, I checked for parenthesis.
I did find parenthesis, however, they do not have any expressions inside of
them, meaning that these parenthesis are for multiplying, and not for stating
order. So, you skip parenthesis. Next, you check for exponents, which you
find none of, so you skip over that. Now, we get to multiplying/dividing, so
you multiply the 15 and the 20 to get 300, and the 25 and 3 to get 75.
***Step 3:
65+300+75 --> 440
Now, we get to addition. You simply add everything up to get your final
answer: $440.

Part C:
If Jaxon had $75, then yes, the coefficients would change.
By subtracting $65 from $75, we can see that the total amount of money
from Jaxon's deposits and his lawn-mowing money is $10. Jaxon already
deposits $15 a week, meaning that, while using the current equation, Jaxon
CANNOT have $75 in his bank account. We can change the equation
so that Jaxon is able to have $75 in his savings account. You can change
the coefficient of 15 to 10, and the other coefficient of 25 to 0.
Now Jaxon is able to have $75 in his savings account.

8 0

3 years ago

The students at Midtown Middle school sold flowers as a fundraiser in September and October. In October, they charged $1.50 for

lions [1.4K]

In the question it is given that,

The students at Midtown Middle school sold flowers as a fundraiser in September and October. In October, they charged $1.50 for each flower

And the October price was a 20% increase of the September price.

Let the price of the flower in October be $x

So we have

$x+ \frac{20}{100}x = 1.50 \\ \frac{120x}{100} = 1.50 \\ x = \frac{1.50*100}{120} = 1.25$

So the price of the flower in Septembet is $1.25 .

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3 years ago

Identify the property:

Lorico [155]

Answer:1) commutative property of addition

Step-by-step explanation:

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3 years ago

Bethany has 43 apples.write a subtraction story about the apples that would require regrouping.

RoseWind [281]

Hey there!

21 apples were rotten and got thrown out. The remaining apples were divided equally among 11 people

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3 years ago

Read 2 more answers

There are two games involving flipping a coin. In the first game you win a prize if you can throw between 45% and 55% heads. In

Nina [5.8K]

Answer:

d) 300 times for the first game and 30 times for the second

Step-by-step explanation:

We start by noting that the coin is fair and the flip of a coin has a probability of 0.5 of getting heads.

As the coin is flipped more than one time and calculated the proportion, we have to use the <em>sampling distribution of the sampling proportions</em>.

The mean and standard deviation of this sampling distribution is:

$\mu_p=p\\\\ \sigma_p=\sqrt{\dfrac{p(1-p)}{N}}$

We will perform an analyisis for the first game, where we win the game if the proportion is between 45% and 55%.

The probability of getting a proportion within this interval can be calculated as:

$P(0.45$

referring the z values to the z-score of the standard normal distirbution.

We can calculate this values of z as:

$z_H=\dfrac{p_H-\mu_p}{\sigma_p}=\dfrac{(p_H-p)}{\sqrt{\dfrac{p(1-p)}{N}}}=\sqrt{\dfrac{N}{p(1-p)}}*(p_H-p)>0\\\\\\z_L=\dfrac{p_L-\mu_p}{\sigma_p}=\dfrac{p_L-p}{\sqrt{\dfrac{p(1-p)}{N}}}=\sqrt{\dfrac{N}{p(1-p)}}*(p_L-p)$

If we take into account the z values, we notice that the interval increases with the number of trials, and so does the probability of getting a value within this interval.

With this information, our chances of winning increase with the number of trials. We prefer for this game the option of 300 games.

For the second game, we win if we get a proportion over 80%.

The probability of winning is:

$P(p>0.8)=P(z>z^*)$

The z value is calculated as before:

$z^*=\dfrac{p^*-\mu_p}{\sigma_p}=\dfrac{p^*-p}{\sqrt{\dfrac{p(1-p)}{N}}}=\sqrt{\dfrac{N}{p(1-p)}}*(p^*-p)>0$

As (p*-p)=0.8-0.5=0.3>0, the value z* increase with the number of trials (N).

If our chances of winnings depend on P(z>z*), they become lower as z* increases.

Then, we can conclude that our chances of winning decrease with the increase of the number of trials.

We prefer the option of 30 trials for this game.

8 0

3 years ago