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2 years ago

Cody has $700 in a savings account that pays 4% simple interest. How much will he have in 1 year?

Mathematics

1 answer:

ladessa [460]2 years ago

3 0

Answer:

$ 728

Step-by-step explanation:

Amount of Money Cody has = $700

Simple Interest = 4%

As it is know that interest is always given on annually basis so

Amount of profit he have in one year = Money he have right now * interest

=700 * 4%

= $700 * \frac{4}{100}$

= $ 28

After one year he will have = total money at that time + Profit

= 700 + 28

= $ 728

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If sin theta = -0.7660, which of the following represents an approximate value of tan theta, for 180 degrees < theta < 270

Elis [28]

We have that
sin ∅=-0.7660

the sin ∅ is negative
so
∅ belong to the III or IV quadrant
but
180°< ∅ < 270°
hence
∅ belong to the III quadrant
sin ∅=-0.7660
sin² ∅+cos² ∅=1
cos² ∅=1-sin² ∅------> cos² ∅=1-(0.7660)²-----> 0.4132
cos ∅=√0.4132-----> cos ∅=0.6428
the value of cos ∅ is negative-------> III quadrant
cos ∅=-0.6428

tan ∅=sin ∅/ cos ∅----> tan ∅=-0.7660/-0.6428----> tan ∅=1.1917

the answer is
tan ∅ is 1.1917

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3 years ago

As an estimation we are told £3 is 4 euro.Convert £13.50 to euros

bixtya [17]

Answer:

15.07 Euro

Step-by-step explanation:

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3 years ago

Please help me answer this. I think that for the first one the answer is A, but I am not sure. There are two parts to the questi

Anika [276]

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For the second, the magnitude is the square root of the sum of the squares of the components. This is equal to $\sqrt{(-14)^2+(-21)^2} = \sqrt{637}=7\sqrt{13}$ .

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3 years ago

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The n candidates for a job have been ranked 1, 2, 3,…, n. Let X 5 the rank of a randomly selected candidate, so that X has pmf p

Anon25 [30]

Answer:

A. E(x) = 1/n×n(n+1)/2

B. E(x²) = 1/n

Step-by-step explanation:

The n candidates for a job have been ranked 1,2,3....n. Let x be the rank of a randomly selected candidate. Therefore, the PMF of X is given as

P(x) = {1/n, x = 1,2...n}

Therefore,

Expectation of X

E(x) = summation {xP(×)}

= summation {X×1/n}

= 1/n summation{x}

= 1/n×n(n+1)/2

= n+1/2

Thus, E(x) = 1/n×n(n+1)/2

Value of E(x²)

E(x²) = summation {x²P(×)}

= summation{x²×1/n}

= 1/n

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3 years ago

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hodyreva [135]

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I hope it helps, Regards.

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3 years ago

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