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Svet_ta [14]
2 years ago
12

In the last basketball game, the panthers scored 63 points. This was seven times the number of points that Reilly scored. Write

and solve an equation to find the number of points Reilly scored
Mathematics
2 answers:
tester [92]2 years ago
5 0
9
In this question all you have to do is divide 63 by 7.
sweet-ann [11.9K]2 years ago
5 0
You would first have to do the opposite of seven times which would be division, this would lead to being 63÷7, with this your answer would be 9, you can also check your work by doing 7×9 and getting 63.
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BRAINLIEST if correct
Mila [183]

Answer:

A=38.88 or 38.9

Step-by-step explanation:

If both are similar then the 9 and the 7.2 are the same

9-7.2= 1.8

9x6=54

7.2-1.8=5.4

7.2x5.4=38.88 or 38.9

4 0
2 years ago
Please help me for the brainliest answer
Zinaida [17]

Answer:

5n^2

Step-by-step explanation:

This is a quadratic sequence.

The nth term must have n^2.

The sequence has a factor of 5.

5n^2

5(1)^2=5\\5(2)^2=20\\5(3)^2=45\\5(4)^2=80\\5(5)^2=125

Few terms to this sequence are:

5, 20, 45, 80, 125, 180, 245, 320, 405, 500, 605, 720, 845, 980, 1125, 1280, 1445, 1620, 1805, 2000, 2205, 2420, 2645, 2880, 3125, 3380, 3645, 3920, 4205, 4500, 4805, 5120, 5445, 5780, ...

3 0
3 years ago
Jill’s bowling scores are approximately normally distributed with mean 170 and standard deviation 20, while Jack’s scores are ap
miss Akunina [59]

Answer:

a) The probability of Jack scoring higher is 0.3446

b) They probability of them scoring above 350 is 0.2119

Step-by-step explanation:

Lets call X the random variable that determines Jill's bowling score and Y the random variable that determines jack's. We have

X \simeq N(170,400)\\Y \simeq N(160,225)

Note that we are considering the variance on the second entry, the square of the standard deviation.

If we have two independent Normal distributed random variables, then their sum is also normally distributed. If fact, we have this formulas:

N(\lambda_1, \sigma^2_1) + N(\lambda_2, \sigma^2_2) = N(\lambda_1 + \lambda_2,\sigma^2_1 + \sigma^2_2) \\r* N(\lambda_1, \sigma^2_1) = N(r\lambda_1,r^2\sigma^2_1)  

for independent distributions N(\lambda_1, \sigma^2_1) , N(\lambda_2, \sigma^2_2) , and a real number r.

a) We define Z to be Y-X. We want to know the probability of Z being greater than 0. We have

Z = Y-X = N(160,225) - N(170,400) = N(160,225) + (N(-170,(-1)^2 * 400) = N(-10,625)

So Z is a normal random variable with mean equal to -10 and vriance equal to 625. The standard deviation of Z is √625 = 25.

Lets work with the standarization of Z, which we will call W. W = (Z-\mu)/\sigma = (Z+10)/25. W has Normal distribution with mean 0 and standard deviation 1. We have

P(Z > 0) = P( (Z+10)/25 > (0+10)/25) = P(W > 0.4)

To calculate that, we will use the <em>known </em>values of the cummulative distribution function Φ of the standard normal distribution. For a real number k, P(W < k) = Φ(k). You can find those values in the Pdf I appended below.

Since Φ is a cummulative distribution function, we have P(W > 0.4) = 1- Φ(0.4)

That value of Φ(0.4) can be obtained by looking at the table, it is 0.6554. Therefore P(W > 0.4) = 1-0.6554 = 0.3446

As a result, The probability of Jack's score being higher is 0.3446. As you may expect, since Jack is expected to score less that Jill, the probability of him scoring higher is lesser than 0.5.

b) Now we define Z to be X+Y Since X and Y are independent Normal variables with mean 160 and 170 respectively, then Z has mean 330. And the variance of Z is equal to the sum of the variances of X and Y, that is, 625. Hence Z is Normally distributed with mean 330 and standard deviation rqual to 25 (the square root of 625).

We want to know the probability of Z being greater that 350, for that we standarized Z. We call W the standarization. W is s standard normal distributed random variable, and it is obtained from Z by removing its mean 330 and dividing by its standard deviation 25.

P(Z > 350) = P((Z  - 330)/25 > (350-330)/25) = P(W > 0.8) = 1-Φ(0.8)

The last equality comes from the fact that Φ is a cummulative distribution function. The value of Φ(0.8) by looking at the table is 0.7881, therefore P(X+Y > 350) = 1 - Φ(0.8) = 0.2119.

As you may expect, this probability is pretty low because the mean value of the sum of their combined scores is quite below 350.

I hope this works for you!

Download pdf
6 0
3 years ago
A postage stamp has a area 1160mm^2. About how long is each side ?
Savatey [412]
Each side is about 290mm long
4 0
3 years ago
A rhombus has coordinates A(-6, 3), B(-4, 4), C(-2, 3), and D(-4, 2). What are the coordinates of rhombus A′B′C′D′ after a 90° c
Liula [17]

We are given that the coordinates of the vertices of the rhombus are:
<span><span>A(-6, 3)
B(-4, 4)
C(-2, 3)
D(-4, 2)

To solve this problem, we must plot this on a graphing paper or graphing calculator to clearly see the movement of the graph. If we transform this by doing a counterclockwise rotation, then the result would be:
</span>A(-6, -3)</span>
B(-4, -4)
C(-2, -3)
D(-4, -2)

 

And the final transformation is translation by 3 units left and 2 units down. This can still be clearly solved by actually graphing the plot. The result of this transformation would be:

<span>A′(6, -8)
B′(7, -6)
C′(6, -4)
D′(5, -6)</span>

7 0
3 years ago
Read 2 more answers
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