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Alex73 [517]
3 years ago
13

If andres probability is 43 /60 and mai looks in the bag and sees that there are only 6 blocks in the bag . Should Adre change h

is estimate based on this information if so what should the new estimate be ? If not explain your reasoning
Mathematics
1 answer:
kari74 [83]3 years ago
4 0

Andre picks a block out of a bag 60 times and notes that 43 of them were green.

What should Andre estimate for the probability of picking out a green block from this bag?

Mai looks in the bag and sees that there are 6 blocks in the bag. Should Andre change his estimate based on this information? If so, what should the new estimate be? If not, explain your reasoning.

The probability of picking out a green block is 43/60.

Andre shall not change his estimate based on this information because only the number of blocks left are told but the color of the blocks are still unknown. If those blocks are not green, then there is no need to change the estimate.

<u>Explanation:</u>

Probability is a numerical portrayal of how likely an occasion is to happen or how likely it is that a recommendation is valid. Likelihood is a number somewhere in the range of 0 and 1, where, generally, 0 demonstrates inconceivability and 1 shows sureness.

To calculate the probability, divide the number of events with the possible outcomes. This will give us the likelihood of a solitary occasion happening. On account of rolling a 3 on a bite the dust, the quantity of occasions is 1 (there's just a solitary 3 on each kick the bucket), and the quantity of results is 6.

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Find the equation of the sphere centered at (-9,9, -9) with radius 5. Normalize your equations so that the coefficient of x2 is
olganol [36]
<h2><u>Answer</u>:</h2>

(a) x² + y² + z² + 18(x - y + z) + 218 = 0

(b) (x + 9)² + (y - 9)² + 56 = 0

<h2><u>Step-by-step explanation:</u></h2>

<em>The general equation of a sphere of radius r and centered at C = (x₀, y₀, z₀) is given by;</em>

(x - x₀)² + (y - y₀)² + (z - z₀)² = r²               ------------------(i)

<em>From the question:</em>

The sphere is centered at C = (x₀, y₀, z₀) = (-9, 9, -9) and has a radius r = 5.

<em>Therefore, to get the equation of the sphere, substitute these values into equation (i) as follows;</em>

(x - (-9))² + (y - 9)² + (z - (-9))² = 5²

(x + 9)² + (y - 9)² + (z + 9)² = 25      ------------------(ii)

<em>Open the brackets and have the following:</em>

(x + 9)² + (y - 9)² + (z + 9)² = 25

(x² + 18x + 81) + (y² - 18y + 81) + (z² + 18z + 81) = 25

x² + 18x + 81 + y² - 18y + 81 + z² + 18z + 81 = 25

x² + y² + z² + 18(x - y + z) + 243 = 25

x² + y² + z² + 18(x - y + z) + 218 = 0    [<em>equation has already been normalized since the coefficient of x² is 1</em>]

<em>Therefore, the equation of the sphere centered at (-9,9, -9) with radius 5 is:</em>

x² + y² + z² + 18(x - y + z) + 218 = 0

(2)  To get the equation when the sphere intersects a plane z = 0, we substitute z = 0 in equation (ii) as follows;

(x + 9)² + (y - 9)² + (0 + 9)² = 25

(x + 9)² + (y - 9)² + (9)² = 25

(x + 9)² + (y - 9)² + 81 = 25        [<em>subtract 25 from both sides</em>]

(x + 9)² + (y - 9)² + 81 - 25 = 25 - 25

(x + 9)² + (y - 9)² + 56 = 0

The equation is therefore, (x + 9)² + (y - 9)² + 56 = 0

6 0
3 years ago
The Candela brothers own two pizza restaurants, one on Park Street and one on Bridge Road.
koban [17]

The mean, median and mode are measures of central tendency, that is they tend to indicate the location middle of the data

Required values;

(a) The performance for the week for Park Street

  • Revenue is <u>Q₂ < $7,500 < Q₃</u>
  • The sales for the week is better than <u>72.91%</u> of all sales

The performance for the week for Bridge Road

  • Revenue; <u>Q₂ < $7,100 < Q₃</u>
  • The sale for the week is better than <u>59.87%</u> of all sales

(b) The mean is <u>$3611</u>

The median is $<u>3,600</u>

The standard deviation is $<u>3250</u>

The Interquartile range is $<u>6075</u>

Reason:

The table of values that maybe used to find a solution to the question is given as follows;

\begin{array}{|l|l|l|}\mathbf{Variable} &\mathbf{Park}&\mathbf{Bridge}\\N&36&40\\Mean&6611&5989\\SE \ Mean&597&299\\StDev&3580&1794\\Minimum&800&1800\\Q_1&3600&5225\\Median&6600&6000\\Q_3&9675&7625\\Maximum&14100&8600\end{array}\right]

(a) Park Street revenue = $7,500

Bridge Road's revenue = $7,100

The two stores sold close to but below the 75th percentile

Bridge Road revenue;

The z-score is given as follows;

Z = \dfrac{x - \mu }{\sigma }

  • Z = \dfrac{7100 - 5,989 }{1794 } \approx 0.6193

From the Z-Table, we have;

The percentile= 0.7291

  • Therefore, the sale for the week for Park Street is better than <u>72.91%</u> of all the sales

Park Street revenue;

The z-score is given as follows;

  • Z = \dfrac{7500 - 6611}{3580} \approx 0.25

From the Z-Table, we have;

The percentile = <u>0.5987</u>

  • Therefore, the sale for the week is better than <u>59.87 %</u> of all the sales

(b) Given that the operating cost is $3,000, frim which we have;

The subtracted value is subtracted from the mean and median to find the new value

Profit = The revenue - Cost

New mean = 6611 - 3000 = 3611

  • The new mean = <u>$3,611</u>

The new median = 6600 - 3000 = 3600

  • The new median = <u>$3,600</u>

The standard deviation and the interquartile range remain the same, therefore, we have;

  • The standard deviation = <u>$3,580</u>

The interquartile range = 9675 - 3600 = 6075

  • The interquartile range = <u>6075</u>

Learn more here:

brainly.com/question/21133077

brainly.com/question/23305909

5 0
2 years ago
Jim’s football team started a play on the 50 yard line. On the first play, the team had a 7-yard gain. On the second play, they
lawyer [7]

This is a basic addition/subtraction question. It's saying that they started at the 50 yard line. On the play, they gained 7 yards. So, we'll add 50+7 = 57.

They're at the 57 yard line at this point. On the second play they lost 10 yards (how unfortunate). Thereby, we will subtract 57-10=47.

50 yards +7 yards. -10 yards.

The answer will be that they'll be on the 47 yard line on the next play.

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3 years ago
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Which counterexample shows that the conjecture "All animals that can fly are birds" is false?
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It seems like you forgot to post the answer choices. However, bats are one animal where they can fly but they aren't birds. So this is one counter-example to prove the claim false. Another example would be any flying insects. 
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What is the sum? (1/x+2)+(1/x+3)+(1/X^2+5+6)
PolarNik [594]
For this case we have the following expression:
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 Rewriting we have:
 (1 / x + 2) + (1 / x + 3) + (1 / ((x + 2) * (x + 3)))
 By doing common factor we have:
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 Rewriting:
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