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Alina [70]
2 years ago
10

Felica starts with y coins in her collection. She loses 20 coins, taking her collection to 28. Choose the equation that correctl

y represents this scenario.
Mathematics
1 answer:
Assoli18 [71]2 years ago
4 0

Answer:

The equation representing the scenario is y-20=28.

Step-by-step explanation:

Given:

Number of coins in Felica collections in start = y \ coins

Number of coins she looses = 20\ coins.

Number of coins left in the collection after loosing = 28 \ coins

We need to write the equations representing the scenario.

Solution:

Now we can say that;

Number of coins in Felica collections in start minus Number of coins she looses is equal to Number of coins left in the collection after loosing.

framing in equation form we get;

y-20=28

Hence The equation representing the scenario is y-20=28.

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Step-by-step explanation:

.5, 3.7416, 4

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A small​ plane's maximum takeoff weight is 2000 pounds or less. six passengers weigh an average of 150 pounds each. use an inequ
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What values of c and d make the equation true?
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Answer:

Third option.

Step-by-step explanation:

You need to cube both sides of the equation. Remember the Power of a power property:

(a^m)^n=a^{mn

\sqrt[3]{162x^cy^5}=3x^2y(\sqrt[3]{6y^d})\\\\(\sqrt[3]{162x^cy^5})^3=(3x^2y(\sqrt[3]{6y^d}))^3\\\\162x^cy^5=27x^6y^36y^d

According to the Product of powers property:

(a^m)(a^n)=a^{(m+n)

Then. simplifying you get:

162x^cy^5=162x^6y^{(3+d)}

Now you need to compare the exponents. You can observe that the exponent of "x" on the right side is 6,  then the exponent of "x" on the left side must be 6. Therefore:

c=6

You can notice that the exponent of "y" on the left side is 5,  then the exponent of "x" on the left side must be 5 too. Therefore "d" is:

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What would be the first step in solving this problem -18 = 2x
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If bolt thread length is normally distributed, what is the probability that the thread length of a randomly selected bolt is Wit
KatRina [158]

Answer:

a) 0.5762

b) 0.0214

c) 0.2718

Step-by-step explanation:

It is given that lengths of the bolt thread are normally distributed. So in order to find the required probability we can use the concept of z distribution and z scores.

Part a) Probability that length is within 0.8 SDs of the mean

We have to calculate the probability that the length of a bolt thread is within 0.8 standard deviations of the mean. Recall that a z- score tells us that how many standard deviations away a value is from the mean. So, indirectly we are given the z-scores here.

Within 0.8 SDs of the mean, means from a score of -0.8  to +0.8. i.e. we have to calculate:

P(-0.8 < z < 0.8)

We can find these values from the z table.

P(-0.8 < z < 0.8) = P(z < 0.8) - P(z < -0.8)

= 0.7881 - 0.2119

= 0.5762

Thus, the probability that the thread length of a randomly selected bolt is within 0.8 SDs of its mean value is 0.5762

Part b) Probability that length is farther than 2.3 SDs from the mean

As mentioned in previous part, 2.3 SDs means a z-score of 2.3.

2.3 Standard Deviations farther from the mean, means the probability that z scores is lesser than - 2.3 or greater than 2.3

i.e. we have to calculate:

P(z < -2.3 or z > 2.3)

According to the symmetry rules of z-distribution:

P(z < -2.3 or z > 2.3) = 1 - P(-2.3 < z < 2.3)

We can calculate P(-2.3 < z < 2.3) from the z-table, which comes out to be 0.9786. So,

P(z < -2.3 or z > 2.3) = 1 - 0.9786

= 0.0214

Thus, the probability that a bolt length is 2.3 SDs farther from the mean is 0.0214

Part c) Probability that length is between 1 and 2 SDs from the mean value

Between 1 and 2 SDs from the mean value can occur both above the mean and below the mean.

For above the mean: between 1 and 2 SDs means between the z scores 1 and 2

For below the mean: between 1 and 2 SDs means between the z scores -2 and -1

i.e. we have to find:

P( 1 < z < 2) + P(-2 < z < -1)

According to the symmetry rules of z distribution:

P( 1 < z < 2) + P(-2 < z < -1) = 2P(1 < z < 2)

We can calculate P(1 < z < 2) from the z tables, which comes out to be: 0.1359

So,

P( 1 < z < 2) + P(-2 < z < -1) = 2 x 0.1359

= 0.2718

Thus, the probability that the bolt length is between 1 and 2 SDs from its mean value is 0.2718

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3 years ago
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