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Pani-rosa [81]
3 years ago
14

3/5x5/4 what the awnser? (HURRY PLZ I NEED HELP)

Mathematics
2 answers:
GalinKa [24]3 years ago
6 0

\frac{3}{5} \times \frac{5}{4} \\

Multiply the numerators and the denominators

\frac{3\times5}{5\times4}

To get

\frac{15}{20}

Now simplify, final answer is \frac{3}{4}. (or 0.75 in decimal form)

Ann [662]3 years ago
4 0

Answer:

0.75

Step-by-step explanation:

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Samantha is wrapping a present for her friend's birthday. The length of the box is 12 inches, width is 7 inches, and the height
loris [4]

Answer:

358 inches^2

Step-by-step explanation:

SA = 2(lw + lh + wh)

SA = 2 (12*7 + 12*5 + 7*5)

SA = 358

Hope this is helpful !!!!!

4 0
3 years ago
Can someone please help me?
Nesterboy [21]

Answer:

2

Step-by-step explanation:

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8 0
3 years ago
The number of texts per day by students in a class is normally distributed with a 
kobusy [5.1K]

Answer:

1, 2, 6

Step-by-step explanation:

The z score shows by how many standard deviations the raw score is above or below the mean. The z score is given by:

z=\frac{x-\mu}{\sigma} \\\\where\ x=raw\ score,\mu=mean, \ \sigma=standard\ deviation

Given that mean (μ) = 130 texts, standard deviation (σ) = 20 texts

1) For x < 90:

z=\frac{x-\mu}{\sigma} \\\\z=\frac{90-130}{20} =-2

From the normal distribution table, P(x < 90) = P(z < -2) = 0.0228 = 2.28%

Option 1 is correct

2) For x > 130:

z=\frac{x-\mu}{\sigma} \\\\z=\frac{130-130}{20} =0

From the normal distribution table, P(x > 130) = P(z > 0) = 1 - P(z < 0) = 1 - 0.5 = 50%

Option 2 is correct

3) For x > 190:

z=\frac{x-\mu}{\sigma} \\\\z=\frac{190-130}{20} =3

From the normal distribution table, P(x > 3) = P(z > 3) = 1 - P(z < 3) = 1 - 0.9987 = 0.0013 = 0.13%

Option 3 is incorrect

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z=\frac{x-\mu}{\sigma} \\\\z=\frac{130-130}{20} =0

For x > 100:

z=\frac{x-\mu}{\sigma} \\\\z=\frac{100-130}{20} =-1.5

From the normal table, P(100 < x < 130) = P(-1.5 < z < 0) = P(z < 0) - P(z < 1.5) = 0.5 - 0.0668 = 0.9332 = 93.32%

Option 4 is incorrect

5)  For x = 130:

z=\frac{x-\mu}{\sigma} \\\\z=\frac{130-130}{20} =0

Option 5 is incorrect

6)  For x = 130:

z=\frac{x-\mu}{\sigma} \\\\z=\frac{160-130}{20} =1.5

Since 1.5 is between 1 and 2, option 6 is correct

5 0
3 years ago
Expand the following using the Binomial Theorem and Pascal’s triangle. (x + 2)6 (x − 4)4 (2x + 3)5 (2x − 3y)4 In the expansion o
ivolga24 [154]
\bf (2x+3)^5\implies &#10;\begin{array}{llll}&#10;term&coefficient&value\\&#10;-----&-----&-----\\&#10;1&&(2x)^5(+3)^0\\&#10;2&+5&(2x)^4(+3)^1\\&#10;3&+10&(2x)^3(+3)^2\\&#10;4&+10&(2x)^2(+3)^3\\&#10;5&+5&(2x)^1(+3)^4\\&#10;6&+1&(2x)^0(+3)^5&#10;\end{array}

as you can see, the terms exponents, for the first term, starts at highest, 5 in this case, then every element it goes down by 1, till it gets to 0

for the second term, starts at 0, and every element it goes up by 1, till it gets to the highest

now, to get the coefficient, they way I get it, is "the product of the current coefficient and the exponent of the first term, divided by the exponent of the second term plus 1"

notice the first coefficient is always 1

so...how did we get 10 for the 3rd element?  well, 5*4/2
how did we get 10 for the fourth element?  well, 10*2/4


\bf (2x-3y)^4\implies &#10;\begin{array}{llll}&#10;term&coefficient&value\\&#10;-----&-----&-----\\&#10;1&&(2x)^4(-3y)^0\\&#10;2&+4&(2x)^3(-3y)^1\\&#10;3&+6&(2x)^2(-3y)^2\\&#10;4&+4&(2x)^1(-3y)^3\\&#10;5&+1&(2x)^0(-3y)^4&#10;\end{array}


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and from there, you can simplify the elements of the expansion by combining the coefficients

like for example, the 7th element of (3a+4b)⁸ will then be 1032192a²b⁶
7 0
3 years ago
Read 2 more answers
What is the volume (in cubic units) of a sphere with a radius of 6 units? Assume that π = 3.14 and round your answer to the near
abruzzese [7]

Answer: 904.32  cubic units.

Step-by-step explanation:

The formula for a sphere's volume is V = V = \frac{4}{3}\pi r^{3} \\

Substitute pi for 3.14 and radius for 6. 6^3 = 216

\\V = \frac{4}{3}\ (3.14)( 216) \\\\

Multiply them together, the answer is 904.32  cubic units.

8 0
3 years ago
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