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photoshop1234 [79]

3 years ago

14

PLEASE HELP. i’ll give brainliest if you want!

1 answer:

ira [324]3 years ago

3 0

Answer:

sharing of simple moral stories

Step-by-step explanation:

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student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te

alexdok [17]

Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = $\frac{1}{4}$ .

Thus, the probability that the student does not receive version A = $1-\frac{1}{4}$ = $\frac{3}{4}$ .

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

$\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}$

= $(\frac{1}{4})^3\times (\frac{3}{4})^2+(\frac{1}{4})^4\times (\frac{3}{4})+(\frac{1}{4})^5$

= $(\frac{1}{4})^3\times (\frac{3}{4})[\frac{3}{4}+\frac{1}{4}+(\frac{1}{4})^2]$

= $(\frac{3}{4^4})[1+\frac{1}{16}]$

= $(\frac{3}{256})[\frac{17}{16}]$

= 0.01171875 × 1.0625

= 0.01245

Thus, the probability that at least 3 out of 5 students receive version A is 0.0124

So, in percent the probability is 0.0124 × 100 = 1.24%

To the nearest tenth, the required probability is 1.2%.

4 0

3 years ago

Cody pays $657 for six months of guitar lessons. He pays the same amount for lessons each month. Which of the following is the b

stiv31 [10]

The answer is D. $110

4 0

3 years ago

Plzz help me plzz! I am timed

Sunny_sXe [5.5K]

Answer:

The second answer, and fifth answer

Step-by-step explanation:

7 0

3 years ago

Use the identity x3+y3+z3−3xyz=(x+y+z)(x2+y2+z2−xy−yz−zx) to determine the value of the sum of three integers given: the sum of

fenix001 [56]

ffn

$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$

$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-(xy+yz+zx))$

the sum of their squares is 110, So $x^2+y^2 + z^2= 110$

the sum of their cubes is 684, so $x^3+y^3 + z^3= 684$

the product of the three integers is 210, so xyz= 210

the sum of any two products (xy+yz+zx) is 107

Now we plug in all the values in the identity

$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-(xy+yz+zx))$

684 - 3(210) = (x+y+z)(110-107)

684 - 630 = (x+y+z)(3)

54 = 3(x+y+z)

Divide by 3 on both sides

18 = x+y+z

the value of the sum of three integers is 18

3 0

3 years ago

Which number is the smallest? A) 1.28 x 10-4 Eliminate B) 2.85 x 10-6 C) 3.24 x 10-5 D) 5.48 x 10-8

vfiekz [6]

The answer is D 5.48*10-8

hope this helps

6 0

3 years ago

Read 2 more answers