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3 years ago

Pls need help desperately don’t know what to write for my essay share ideas

Mathematics

2 answers:

Dafna1 [17]3 years ago

4 0

Write about the police situations that’s going on right now

SCORPION-xisa [38]3 years ago

3 0

According to the picture you can write on success

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Find the a26 arithmetic sequence: -15,-35,-55,-75 -515 460 -490

Vlad [161]

Step-by-step explanation:

<h3>QUESTION:-</h3>

Find 26th term of the A.P --15,-35,-55,-75...

<h3>Solution:-</h3>

First term=a=-15

Common Difference=d=-20

We know that

$\boxed{\sf a_n=a+(n-1)d}$

$\\ \:\:\:\:\sf {:}\dashrightarrow a_{26}=-15+(26-1)(-20)$

$\\ \:\:\:\:\sf {:}\dashrightarrow a_{26}=-15+25(-20)$

$\\ \:\:\:\:\sf {:}\dashrightarrow a_{26}=-15+(-500)$

$\\ \:\:\:\:\sf {:}\dashrightarrow a_{26}=-15-500$

$\\ \:\:\:\:\sf {:}\dashrightarrow {\boxed{\bf a_{26}=-515}}$

6 0

3 years ago

The third graders collected 754 cans the fourth graders collected 592 cans estimate how many more cans the third graders collect

MakcuM [25]

754 minus 592 is 162 so estimated to the tens would be 160.

8 0

2 years ago

Read 2 more answers

Someone Help please!

salantis [7]

I think ur answer would be C.

8 0

2 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 known 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

Please help ( the function table)

Norma-Jean [14]

Hope this will help

8 0

3 years ago