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Oliga [24]
2 years ago
6

Find the value of P such that 2 ^p sqrt 4 x 1/4

Mathematics
1 answer:
Sav [38]2 years ago
5 0

Answer: p=-1

Step-by-step explanation:

To find the value of p, we can manipulate the right side so that it looks like the left.

We know that \sqrt{4}=2. Therefore, it can be written as 2¹.

\frac{1}{4} is the same as \frac{1}{2^2 } because 2²=4. To make it not a fraction, we can flip it and change the exponent. \frac{1}{4}=2^-^2

Now, we can put them together.

2^p=2^1*2^-^2         [multiply]

2^p=2^-^1

Now, we know that p=-1.

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Make y the subject: <br><br> 5x + 2y = 11
Anton [14]
Does that mean solve for y? If so, here it is:
5x + 2y = 11.

Move the x term by subtracting.
2y = -5x + 11

Divide by 2.
y = -5/2x + 11/2
7 0
3 years ago
Read 2 more answers
If an event has a 55% chance of happening in one trial, how do I determine the chances of it happening more than once in 4 trial
aev [14]

The chances of it happening more than once in 4 trials is 13%

<h3>How to determine the number</h3>

From the information given, we have can deduce that;

Probability of 1 trial = 55%

= 55/ 100

Find the ratio

= 0. 55

We are to find the probability of it happening more than once in 4 different trials

If the probability of it happening in one trial is 555 which equals 0. 55

Then the probability of it happening in 1 in 4 trials is given as;

P(1/4 trials) = 1/ 4 × 55%

P(1/4 trials) = 1/ 4 × 0. 55

Put in decimal form

P(1/4 trials) = 0. 25 × 0. 55

P(1/4 trials) = 0. 138

But we have to know the percentage

= 0. 138 × 100

Multiply the values, we have

= 13. 8 %

Thus, the chances of it happening more than once in 4 trials is 13%

Learn more about probability here:

brainly.com/question/24756209

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6 0
2 years ago
Find exact value for cos(pi/16)
stellarik [79]
The exact value for cos(pi/16) would be :

cos (pi/16) =  +/-  √1+cosa/2

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Hope this helps
6 0
3 years ago
Change the line 3x − 8y = 5 into slope-intercept form.
Nimfa-mama [501]
Slope is y=mx+b
m=slope
b=y intercept

solve for y basically


3x-8y=5
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-8y=-3x+5
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4 0
3 years ago
A bag contains 6 red apples and 5 yellow apples. 3 apples are selected at random. Find the probability of selecting 1 red apple
tester [92]

Answer:  The required probability of selecting 1 red apple and 2 yellow apples is 36.36%.

Step-by-step explanation:  We are given that a bag contains 6 red apples and 5 yellow apples out of which 3 apples are selected at random.

We are to find the probability of selecting 1 red apple and 2 yellow apples.

Let S denote the sample space for selecting 3 apples from the bag and let A denote the event of selecting 1 red apple and 2 yellow apples.

Then, we have

n(S)=^{6+5}C_3=^{11}C_3=\dfrac{11!}{3!(11-3)!}=\dfrac{11\times10\times9\times8!}{3\times2\times1\times8!}=165,\\\\\\n(A)\\\\\\=^6C_1\times^5C_2\\\\\\=\dfrac{6!}{1!(6-1)!}\times\dfrac{5!}{2!(5-2)!}\\\\\\=\dfrac{6\times5!}{1\times5!}\times\dfrac{5\times4\times3!}{2\times1\times3!}\\\\\\=6\times5\times2\\\\=60.

Therefore, the probability of event A is given by

P(A)=\dfrac{n(A)}{n(S)}=\dfrac{60}{165}=\dfrac{4}{11}\times100\%=36.36\%.

Thus, the required probability of selecting 1 red apple and 2 yellow apples is 36.36%.

4 0
3 years ago
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