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Tasya [4]
3 years ago
6

4(c+12)=2c+18 what is c=?

Mathematics
1 answer:
Elanso [62]3 years ago
5 0
4(c+12)=2c+18
4c+12c=2c+18
4c-2c=18-48
2c=-30
c= -30/2
c= -15
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Please answer these questions!
ycow [4]

for the first one it’s 39%. Hope this helps

8 0
2 years ago
The random numbers below represent 15 trials of a basketball simulation conducted using a spinner numbered 0–8. 76645, 46757, 28
mote1985 [20]

Answer: Option D.

Step-by-step explanation:

Out of the 15 simulations, the number with a 1 in them is 7. (are the sets of 5 shots that have at least one 3-pointer)

So out of 15 sets, 7 of them have at least one 3-point in them.

the experimental probability is p = 7/15.

So the correct option is D.

4 0
3 years ago
the figure is made of up of a cone and a cylinder, to the nearest whole number what is the approximate volume of this figure. PL
Deffense [45]
The total volume is the sum of volume of cone and volume of cylinder.

Volume of cylinder = πr²h
π = 3.14
r = 4/2 = 2 mm
h = 2 mm

So, volume of cylinder equals:

3.14 *(2)^{2}*2 =25.12 mm³

Volume of cone = \frac{1}{3} \pi  r^{2}h
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h = 3 mm

So, volume of cylinder equals:

\frac{1}{3}*3.14* (2)^{2}*3=18.84mm³

Thus, total volume of the figure = 25.12 + 18.84 = 43.96 mm³

7 0
3 years ago
15x²Y - 10xy^3<br> plz help
nikklg [1K]

Answer:

5 x (3 x Y - 2y^3)

x(15 x Y - 10 y^3)

5 (y^3 - 3x Y )^    -    5y^6

-----------------------------------

3Y                                3Y

X=  0       Y=  0

HOPE THIS HELPSS

3 0
3 years ago
How do you find the normal approximation without using the continuity correction?
Dennis_Churaev [7]
Let's say you want to compute the probability \mathbb P(a\le X\le b) where X converges in distribution to Y, and Y follows a normal distribution. The normal approximation (without the continuity correction) basically involves choosing Y such that its mean and variance are the same as those for X.

Example: If X is binomially distributed with n=100 and p=0.1, then X has mean np=10 and variance np(1-p)=9. So you can approximate a probability in terms of X with a probability in terms of Y:

\mathbb P(a\le X\le b)\approx\mathbb P(a\le Y\le b)=\mathbb P\left(\dfrac{a-10}3\le\dfrac{Y-10}3\le\dfrac{b-10}3\right)=\mathbb P(a^*\le Z\le b^*)

where Z follows the standard normal distribution.
8 0
3 years ago
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