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3 years ago

How to solve 5s+7−2s=16

Mathematics

2 answers:

baherus [9]3 years ago

8 0

5s+7−2s=16

combine like terms

3s + 7 = 16

subtract 7 from both sides

3s = 9

divide both sides by 3

s = 3

answer

s = 3

Rainbow [258]3 years ago

7 0

5s+7-2s=16

(5-2)s=16-7

3s = 9

s= 9/3

s=3

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The curves r1(t) = 2t, t2, t4 and r2(t) = sin t, sin 5t, 2t intersect at the origin. Find their angle of intersection, θ, correc

masya89 [10]

Answer:

Therefore the angle of intersection is $\theta =79.48^\circ$

Step-by-step explanation:

Angle at the intersection point of two carve is the angle of the tangents at that point.

Given,

$r_1(t)=(2t,t^2,t^4)$

and $r_2(t)=(sin t , sin5t, 2t)$

To find the tangent of a carve , we have to differentiate the carve.

$r'_1(t)=(2,2t,4t^3)$

The tangent at (0,0,0) is [ since the intersection point is (0,0,0)]

$r'_1(0)=(2,0,0)$ [ putting t= 0]

$|r'_1(0)|=\sqrt{2^2+0^2+0^2} =2$

Again,

$r'_2(t)=(cos t ,5 cos5t, 2)$

The tangent at (0,0,0) is

$r'_2(0)=(1 ,5, 2)$ [ putting t= 0]

$|r'_1(0)|=\sqrt{1^2+5^2+2^2} =\sqrt{30}$

If θ is angle between tangent, then

$cos \theta =\frac{r'_1(0).r'_2(0)}{|r'_1(0)|.|r'_2(0)|}$

$\Rightarrow cos \theta =\frac{(2,0,0).(1,5,2)}{2.\sqrt{30} }$

$\Rightarrow cos \theta =\frac{2}{2\sqrt{30} }$

$\Rightarrow cos \theta =\frac{1}{\sqrt{30} }$

$\Rightarrow \theta =cos^{-1}\frac{1}{\sqrt{30} }$

$\Rightarrow \theta =79.48^\circ$

Therefore the angle of intersection is $\theta =79.48^\circ$ .

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3 years ago

Using the least number of coins how many quarters, dimes, nickels, and pennies to make $1.74

SpyIntel [72]

Let's use our largest amount (quarter) to see how many can fit evenly into 1.74.

1.74/0.25 = 6.96
The most quarters that can fit into $1.74 is 6.

---Multiply $0.25 by 6 = $1.50

---Subtract $1.50 from $1.74 = $0.24

---We need to use the rest of the coins to fill up $0.24.

Now let's use our second largest amount (dime) to see how many can fit evenly into 0.24.

0.24/0.10 = 2.4
The most dimes that can fit into $0.24 is 2.

---Multiply $0.10 by 2 = $0.20

---Subtract $0.20 from $0.24 = $0.04

No nickels can fit into $0.04, because they are worth $0.05.

We can use our least amount (pennies) to fill in the $0.04 remaining.

Your answer is 6 quarters, 2 dimes, and 4 pennies.

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3 years ago

Chan deposited money into his retirement account that is compounded annually at an interest rate of 7%. Chan thought the equival

dedylja [7]

Answer:

No, equivalent quarterly rate will be approx 1.75%

Step-by-step explanation:

Given that Chan deposited money into his retirement account that is compounded annually at an interest rate of 7%.

We know that there are 4 quarters in 1 year.

So to find that equivalent quarterly we will divide given yearly rate by number of quarters.

That means divide 7% by 4.

which gives 1.75%.

But that is different than Chan's though of 2% quarterly interest.

Hence Chan is wrong.

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3 years ago

When circuit boards used in the manufacture of compact disc players are tested, the long-run percentage of defectives is 5%. Let

sergiy2304 [10]

Answer:

(a) P(X=3) = 0.093

(b) P(X≤3) = 0.966

(c) P(X≥4) = 0.034

(d) P(1≤X≤3) = 0.688

(e) The probability that none of the 25 boards is defective is 0.277.

(f) The expected value and standard deviation of X is 1.25 and 1.089 respectively.

Step-by-step explanation:

We are given that when circuit boards used in the manufacture of compact disc players are tested, the long-run percentage of defectives is 5%.

Let X = the number of defective boards in a random sample of size, n = 25

So, X ∼ Bin(25,0.05)

The probability distribution for the binomial distribution is given by;

$P(X=r)= \binom{n}{r} \times p^{r}\times (1-p)^{n-r} ; x = 0,1,2,......$

where, n = number of trials (samples) taken = 25

r = number of success

p = probability of success which in our question is percentage

of defectivs, i.e. 5%

(a) P(X = 3) = $\binom{25}{3} \times 0.05^{3}\times (1-0.05)^{25-3}$

= $2300 \times 0.05^{3}\times 0.95^{22}$

= 0.093

(b) P(X $\leq$ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

= $\binom{25}{0} \times 0.05^{0}\times (1-0.05)^{25-0}+\binom{25}{1} \times 0.05^{1}\times (1-0.05)^{25-1}+\binom{25}{2} \times 0.05^{2}\times (1-0.05)^{25-2}+\binom{25}{3} \times 0.05^{3}\times (1-0.05)^{25-3}$

= $1 \times 1 \times 0.95^{25}+25 \times 0.05^{1}\times 0.95^{24}+300 \times 0.05^{2}\times 0.95^{23}+2300 \times 0.05^{3}\times 0.95^{22}$

= 0.966

(c) P(X $\geq$ 4) = 1 - P(X < 4) = 1 - P(X $\leq$ 3)

= 1 - 0.966

= 0.034

(d) P(1 ≤ X ≤ 3) = P(X = 1) + P(X = 2) + P(X = 3)

= $\binom{25}{1} \times 0.05^{1}\times (1-0.05)^{25-1}+\binom{25}{2} \times 0.05^{2}\times (1-0.05)^{25-2}+\binom{25}{3} \times 0.05^{3}\times (1-0.05)^{25-3}$