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Dafna1 [17]
3 years ago
7

I got stuck, I don’t even know where to start I think the answer is A but idk. Help pls

Mathematics
1 answer:
kolbaska11 [484]3 years ago
4 0

Answer:

yeah, it's A your correct!

hope it helps;)

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Pls help
denis-greek [22]

Answer:

C I think

Step-by-step explanation:

Hehehehehehe

7 0
3 years ago
a bag contains eleven counters. five are white. a counter is taken out the bag and is not replaced. a second counter is taken ou
Fed [463]

Answer:

\displaystyle P(A)=\frac{6}{11}

Step-by-step explanation:

<u>Probabilities</u>

The question describes an event where two counters are taken out of a bag that originally contains 11 counters, 5 of which are white.

Let's call W to the event of picking a white counter in any of the two extractions, and N when the counter is not white. The sample space of the random experience is

\Omega=\{WW,WN,NW,NN\}

We are required to compute the probability that only one of the counters is white. It means that the favorable options are

A=\{WN,NW\}

Let's calculate both probabilities separately. At first, there are 11 counters, and 5 of them are white. Thus the probability of picking a white counter is

\displaystyle \frac{5}{11}

Once a white counter is out, there are only 4 of them and 10 counters in total. The probability to pick a non-white counter is now

\displaystyle \frac{6}{10}

Thus the option WN has the probability

\displaystyle P(WN)=\frac{5}{11}\cdot \frac{6}{10}=\frac{30}{110}=\frac{3}{11}

Now for the second option NW. The initial probability to pick a non-white counter is

\displaystyle \frac{6}{11}

The probability to pick a white counter is

\displaystyle \frac{5}{10}

Thus the option NW has the probability

\displaystyle P(NW)=\frac{6}{11}\cdot \frac{5}{10}=\frac{30}{110}=\frac{3}{11}

The total probability of event A is the sum of both

\displaystyle P(A)=\frac{3}{11}+\frac{3}{11}=\frac{6}{11}

\boxed{\displaystyle P(A)=\frac{6}{11}}

7 0
2 years ago
MATHS | Cousin needs help❤️ question bellow
Licemer1 [7]
Equation: y = 6 - 2x

Fill in the x numbers in the x which is next to -2 —> y = 6 - 2(0), y = 6 -2(1), y = 6 -2(3), and etc.

A) 0 —> 6
1 —> 4
2 —> 2
3 —> 0
4 —> -2
5 —> -4

B) Put a dot on y axis on number 6, then go down by 2 because your subtracting 2

C) 6 - 2x = 3
-6 -6
-2x = -3
-2/-2 -3/-2

X = 1.5
3 0
3 years ago
Find the volume of the solid under the plane 5x + 9y − z = 0 and above the region bounded by y = x and y = x4.
svp [43]
<span>For the plane, we have z = 5x + 9y

For the region, we first find its boundary curves' points of intersection.
x = x^4 ==> x = 0, 1.

Since x > x^4 for y in [0, 1],

The volume of the solid equals

\int\limits^1_0 { \int\limits_{x^4}^x {(5x+9y)} \, dy } \, dx = \int\limits^1_0 {\left[5xy+ \frac{9}{2} y^2\right]_{x^4}^{x}} \, dx  \\  \\ =\int\limits^1_0 {\left[\left(5x(x)+ \frac{9}{2} (x)^2\right)-\left(5x(x^4)+ \frac{9}{2} (x^4)^2\right)\right]} \, dx  \\  \\ =\int\limits^1_0 {\left(5x^2+ \frac{9}{2} x^2-5x^5- \frac{9}{2} x^8\right)} \, dx =\int\limits^1_0 {\left( \frac{19}{2} x^2-5x^5- \frac{9}{2} x^8\right)} \, dx \\  \\ =\left[ \frac{19}{6} x^3- \frac{5}{6} x^6- \frac{1}{2} x^9\right]^1_0

=\frac{19}{6} - \frac{5}{6} - \frac{1}{2} =\bold{ \frac{11}{6} \ cubic \ units}</span>
8 0
3 years ago
What is 80 percent of 50
svet-max [94.6K]
40 Why? 80%x X=50 or 80/100x X=50
3 0
2 years ago
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