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Fed [463]
3 years ago
13

221 is blank% of 650

Mathematics
1 answer:
Yuri [45]3 years ago
8 0

Answer:

34%

hope this helps

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Find 13% of 400. Enter the correct answer
ANTONII [103]

Answer:

the answer to your question would be 52

Step-by-step explanation:

8 0
3 years ago
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Find f”(x)=(3x^4+2)^5
Sidana [21]

Answer:

180x²(3x⁴+2)⁴+240x³(3x⁴+2)³

Step-by-step explanation:

we need to break this up using the chain rule

(3x⁴+2)=u

so we have

So now we take 3x⁴+2 and take the derivative (which I will assume you know how to do, if not lmk)

which is 12x³

we now need to take the derivative of u⁵=5u⁴

Mulitply these together

60x³u⁴=60x³(3x⁴+2)⁴

do the same thing again

60x³u⁴

we now need to do the mulitplication rleu

(60x³)'(u⁴)+(60x³)(u⁴)'

=

180x²u⁴+60x³*4u³

simplify

180x²(3x⁴+2)⁴+240x³(3x⁴+2)³

while you could do more to simplify it, it would get really messy

4 0
3 years ago
Write an expression to represent the length is 6 feet longer than the width
Ivenika [448]

Step-by-step explanation:

Let l be the length of the room, and w be the width.

If the length is 6 feet longer than twice the width, then we can say l = 2w+6

Knowing that the formula for the perimeter of a rectangle is P = 2l+2w, and knowing that P=144 ft, we can say that 2L + 2w= 144 ft.

To find the dimensions, we plug our value for l into our perimeter equation

This gives us the following equation: 144= 2(2w+6) + 2w, which simplifies to 144=4w+12+2w, which further simplifies to 144= 6w+12

To get w on one side of the equation, we subtract 12 from each side, which gives us 132=6w

Dividing each side by 6, we determine that w= 22 ft.

Plugging this value back into our first equation we see that l= 2(22)+6

So l= 50 ft.

So, the dimensions of the room are as follows: length is 50 ft., width is 22 ft.

5 0
3 years ago
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Jack’s oak tree is 10 feet tall and grows
dalvyx [7]

Answer: 0 months its already shorter than hanks tree

8 0
3 years ago
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The weights of the chocolate in Hershey Kisses are normally distributed with a mean of 4.5338 g and a standard deviation of 0.10
dusya [7]

The distribution is not a standard normal distribution.

We know that the variable (X) usually obeys and follows a standard normal distribution provided that the summation of (X) is zero and the variance of the random variable V(X) = 1.

Mathematically:

E(X) = 0 and V(X) = 1

Given that;

  • the mean which follows a normal distribution (X) = 4.5338 g, and
  • the standard deviation SD(X) = 0.1039 g

As such, the distribution is not a standard normal distribution.

Therefore, we can conclude that the distribution is not a standard normal distribution.

Learn more about standard normal distribution here:

brainly.com/question/11876263?referrer=searchResults

7 0
3 years ago
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