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3 years ago

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A study shows that 70 out of 200 college students read newspapers . marysville college has 12,600 students. based on sturdy , ho

w many students at marysville college are likely to read the newspaper?
A. 3,600
B. 4,410
C. 4,900
D. 8,820

2 answers:

d1i1m1o1n [39]3 years ago

7 0

441
Hope this helps
I

Vikentia [17]3 years ago

7 0

( B ). The answer is b I think if not sorry

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Find the simple interest on a $3,219.00 principal, deposited for six years at a rate of 1.51%.

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Answer:

s.i = P. T. R /100

= 3219 . 6 . 1.51 /100

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3 years ago

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Find the roots of h(t) = (139kt)^2 − 69t + 80

Sonbull [250]

Answer:

The positive value of $k$ will result in exactly one real root is approximately 0.028.

Step-by-step explanation:

Let $h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80$ , roots are those values of $t$ so that $h(t) = 0$ . That is:

$19321\cdot k^{2}\cdot t^{2}-69\cdot t + 80=0$ (1)

Roots are determined analytically by the Quadratic Formula:

$t = \frac{69\pm \sqrt{4761-6182720\cdot k^{2} }}{38642}$

$t = \frac{69}{38642} \pm \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$

The smaller root is $t = \frac{69}{38642} - \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$ , and the larger root is $t = \frac{69}{38642} + \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$ .

$h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80$ has one real root when $\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} = 0$ . Then, we solve the discriminant for $k$ :

$\frac{80\cdot k^{2}}{19321} = \frac{4761}{1493204164}$

$k \approx \pm 0.028$

The positive value of $k$ will result in exactly one real root is approximately 0.028.

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3 years ago

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Alona [7]

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What lump sum do parents need to deposit in an account earning 9%, compounded monthly, so that it will grow to $100,000 for thei

gogolik [260]

Answer:

They need to deposit $31,172.49

Step-by-step explanation:

The compound interest formula is given by:

$A = P(1 + \frac{r}{n})^{nt}$

Where A is the amount of money, P is the principal(the initial sum of money), r is the interest rate(as a decimal value), n is the number of times that interest is compounded per year and t is the time the money is invested or borrowed for, in years.

In this problem

We want to find P for which $A = 100000$ when $r = 0.09, n = 12, t = 13$

So

$A = P(1 + \frac{r}{n})^{nt}$

$100000 = P(1 + \frac{0.09}{12})^{12*13}$

$3.208P = 100000$

$P = \frac{100000}{3.208}$

$P = 31172.49$

They need to deposit $31,172.49

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