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2 years ago

11

Line segment ON is perpendicular to line segment ML. What is the length of segment NP?

1 answer:

boyakko [2]2 years ago

7 0

Answer:

2 units.

Step-by-step explanation:

In this question we use the Pythagorean theorem which is shown below:

Given that

The right triangle OMP

The hypotenuse i.e OM is the circle radius =5 units.

The segment MP = 4 units length

Therefore

$OP^2 + MP^2 = OM^2$

$OP^2 + 4^2 = 5^2$

$OP^2 + 16 = 25$

So OP is 3

Now as we can see that ON is also circle radius so it would be 5 units

And,

ON = OP + PN

So,

PN is

= ON - OP

= 5 units - 3 units

= 2 units

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Chips suppose a computer chip manufacturer rejects 2% of the chips produced because they fail presale testing.

Artyom0805 [142]

(a) P( fifth one is bad) = P( first 4 are OK) * P(5th is bad)

= (0.98)^4 * 0.02 = 0.0184 or 1.84%

(b) this will be (0.98)^10 = 81.70%

4 0

3 years ago

In the expansion (ax+by)^7, the coefficients of the first two terms are 128 and -224, respectively. Find the values of a and b

madam [21]

Answer:

a = 2, b = 3.5

Step-by-step explanation:

Expanding $(ax+by)^7$ using Binomial expansion, we have that:

$(ax+by)^7$ =

$(ax)^7(by)^0 + (ax)^6(by)^1 + (ax)^5(by)^2 + (ax)^4(by)^3 + (ax)^3(by)^4 + (ax)^2(by)^5 + (ax)^1(by)^6 + (ax)^0(by)^7$

$= (a)^7(x)^7+ (a)^6(x)^6(b)(y) + (a)^5(x)^5(b)^2(y)^2 + (a)^4(x)^4(b)^3(y)^3 + (a)^3(x)^3(b)^4(y)^4 + (a)^2(x)^2(b)^5(y)^5 + (a)(x)(b)^6(y)^6 + (b)^7(y)^7\\\\\\= (a)^7(x)^7+ (a)^6(b)(x)^6(y) + (a)^5(b)^2(x)^5(y)^2 + (a)^4(b)^3(x)^4(y)^3 + (a)^3(b)^4(x)^3(y)^4 + (a)^2(b)^5(x)^2(y)^5 + (a)(b)^6(x)(y)^6 + (b)^7(y)^7$

We have that the coefficients of the first two terms are 128 and -224.

For the first term:

=> $a^7 = 128$

=> $a = \sqrt[7]{128}\\ \\\\a = 2$

For the second term:

$a^6b = -224$

$b = \frac{-224}{a^6}$

$b = \frac{-224}{2^6} \\\\\\b = \frac{-224}{64} \\\\\\b = 3.5$

Therefore, a = 2, b = 3.5

5 0

2 years ago

Help me get the answer

drek231 [11]

This is hard- DHJDHSJS

3 0

3 years ago

Two golfers completed one round of golf. The first golfer had a score of +6 and the second golfer had a score of –3. How many mo

Monica [59]

Question 16

Given that two golfers completed one round of golf.

It is stated that the first golfer had a score of +6 and the second golfer had a score of -3.

Let suppose 'x' be the number of shots the first golfer took.

It is clear that the first golfer is 6 over par, while the second golfer is under 3.

Thus, the equation becomes:

$-3 + x = 6$

adding -3 to both sides

$-3+3+x = 6+3$

simplify

$x = 9$

Thus, the second golfer has taken 9 more shorts.

Question 17

When we talk about credit, it means:

Credit = +

When we talk about debt, it means:

Debit = -

Thus,

Credit = +84

Debt = -29

Balance = 84 + (-29)

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Thus, the balance is: 55

6 0

2 years ago

: A new bear population that begins with 150 bears in 2000 decreases at a rate of 20% per year.

d1i1m1o1n [39]

Answer:

2 bears in 2020.

Step-by-step explanation:

We have been given that a new bear population that begins with 150 bears in 2000 decreases at a rate of 20% per year.

We will use exponential decay formula to solve our given problem as:

$y=a\cdot (1-r)^x$ , where,

y = Final quantity,

a = Initial value,

r = Decay rate in decimal form,

x = Time

$20\%=\frac{20}{100}=0.20$

Upon substituting our given values in above formula, we will get:

$y=150(1-0.20)^x$

$y=150(0.80)^x$ , where x corresponds to year 2000.

To find the population in 2020, we will substitute $x=20$ in our equation as:

$y=150(0.80)^{20}$

$y=150(0.011529215046)$

$y=1.7293822569$

$y\approx 2$

Therefore, 2 bears are there predicted to be in 2020.

Since population is decreasing so population is best described as exponential decay.

8 0

3 years ago