All categories

3 years ago

Suppose that each quarter at a particular community college Jared has to pay $729 in tuition and $22 in fees.

Mathematics

1 answer:

k0ka [10]3 years ago

6 0

Answer:

$\$3755$

Step-by-step explanation:

For 1 Quarter:

$Amount\ paid\ in\ tuition=\$729\\\\Amount\ paid\ in\ fees=\$22\\\\Total\ amount\ paid\ in\ 1\ Quarter=Amount\ paid\ in\ tuition+Amount\ paid\ in\ fees\\\\Total\ amount\ paid\ in\ 1\ Quarter=729+22\\\\Total\ amount\ paid\ in\ 1\ Quarter=\$ 751$

For 5 Quarter:

$Remaining\ Quarters=5\\\\Total\ amount\ paid\ in\ 5\ Quarter=Number\ of\ quarters\times Total\ amount\ paid\ in\ 1\ Quarter\\\\Total\ amount\ paid\ in\ 5\ Quarter=5\times 751\\\\Total\ amount\ paid\ in\ 5\ Quarter=\$3755\\\\So\ total\ \$3755\ is\ required\ for\ tuition\ and\ fees.$

You might be interested in

Suppose u = (1,2,3), v = (2, -3, -4), and w = (1,0,-2). (a) Find u-2v + 3w. (b) Write the vector (3,4,5) as a linear combination

marishachu [46]

<h2>Answer with explanation:</h2>

We are given vectors u,v and w as follows:

$u=(1,2,3)\,\ v=(2,-3,-4),\ and\ w=(1,0,-2)$

(a)

Now we are asked to find the value of:

$u-2v+3w$

The value is calculated as follows:

$u-2v+3w=(1,2,3)-2(2,-3,-4)+3(1,0,-2)\\\\i.e.\\\\u-2v+3w=(1,2,3)+(-2\times 2,-2\times -3,-2\times -4)+(3\times 1,3\times 0,3\times -2)\\\\i.e.\\\\u-2v+3w=(1,2,3)+(-4,6,8)+(3,0,-6)\\\\i.e.\\\\u-2v+3w=(1-4+3,2+6+0,3+8-6)\\\\i.e.\\\\u-2v+3w=(0,8,5)$

Hence,

$u-2v+3w=(0,8,5)$

(b)

We are asked to represent (3,4,5) as a linear combination of u, v, and w.

Let:

$(3,4,5)=au+bv+cw$

$(3,4,5)=a(1,2,3)+b(2,-3,-4)+c(1,0,-2)$

i.e.

$(3,4,5)=(a,2a,3a)+(2b,-3b,-4b)+(c,0,-2c)\\\\i.e.\\\\(3,4,5)=(a+2b+c,2a-3b,3a-4b-2c)\\\\i.e.\\\\a+2b+c=3----------(1)$

$2a-3b=4-------------(2)$

$3a-4b-2c=5------------(3)$

on multiplying equation(1) by 2 an then adding equation (1) to equation (3)

$5a=11\\\\i.e.\\\\a=\dfrac{11}{5}$

on putting the value of a in equation (2) we get:

$2\times \dfrac{11}{5}-3b=4\\\\i.e.\\\\\dfrac{22}{5}-3b=4\\\\i.e.\\\\3b=\dfrac{22}{5}-4\\\\i.e.\\\\3b=\dfrac{2}{5}\\\\i.e.\\\\b=\dfrac{2}{15}$

Now on putting the value of a and b in equation (1)

$\dfrac{11}{5}+2\times \dfrac{2}{15}+c=3\\\\i.e.\\\\\dfrac{11}{5}+\dfrac{4}{15}+c=3\\\\i.e.\\\\c=3-\dfrac{11}{5}-\dfrac{4}{15}\\\\i.e.\\\\c=\dfrac{3\times 15-11\times 3-4}{15}\\\\i.e.\\\\c=\dfrac{45-33-4}{15}\\\\i.e.\\\\c=\dfrac{8}{15}$

Hence, we have:

$(3,4,5)=\dfrac{11}{5}(1,2,3)+\dfrac{2}{15}(2,-3,-4)+\dfrac{8}{15}(1,0,-2)\\\\i.e.\\\\(3,4,5)=\dfrac{11}{5}u+\dfrac{2}{15}v+\dfrac{8}{15}w$

5 0

4 years ago

Assume that the weights of individuals are independent and normally distributed with a mean of 160 pounds and a standard deviati

Ilia_Sergeevich [38]

Answer:

a) $\bf 0.3446^{25}=2.7095*10^{-12}$

b) 3,624.25 pounds

Step-by-step explanation:

Since 4300/25 = 172, in average every single person should weight 172 pounds to exceed the design limit.

The probability that one person weights 172 pounds or more is the area under the Normal curve with mean 160 pounds and standard deviation 30 pounds to the right of 172.

In Excel and OpenOffice Calc, this value is found with the formula

=1-NORMDIST(172;160;30;1)

(NORMDIST(172;160;30;1) gives the area to the left of 172, so 1-NORMDIST(172;160;30;1) gives the area to the right of 172)

and equals 0.3446

(see picture 1)

Hence, the probability that all the 25 people exceeds 172 pounds equals

$\bf 0.3446^{25}=2.7095*10^{-12}$

Similarly, we must find a weight w such that if p is the area to the left of w, then

$\bf p^{25}>0.0001$

$\bf p=\sqrt[25]{0.0001}=0.6918$

w would be the point such that the area under the Normal curve with mean 160 pounds and standard deviation 30 pounds to the right of w equals 0.6918.

In Excel and OpenOffice Calc this is found with

=NORMINV(1-0.6918;160;30)

and equals 144.97 pounds

(See picture 2)

and the design limit that is exceeded by 25 occupants with probability 0.0001 is 25*144.97 = 3,624.25 pounds

6 0

4 years ago

An ice cream vendor sells three flavors: chocolate, strawberry, and vanilla. Forty five percent of the sales are chocolate, whil

olganol [36]

Answer:

Step-by-step explanation:

Sales of the chocolate ice cream is 45%, 30% for strawberry and 25% (100%-45%-30%) of vanilla.

Percentages of cone sales for chocolate, strawberry and vanilla are 75%, 60% and 40%respectively.

A. Probability chocolate chosen: 45%=0.45

B. Probability strawberry chosen: 30%=0.30

C. Probability vanilla chosen:25%=0.25

D. Probability ice cream on a cone: 75%×45%+60%×30%+40%×25%

=0.75×0.45+0.60×0.30+0.40×0.25

=0.3375+0.18+0.1

=0.6175

E. Probability ice cream in a cup: 1- Probability ice cream in a cone

=1-0.6175

=0.3825

Probability that the ice cream sold on a cone and was strawberry flavoured is : 30%×60%

=0.30×0.60

=0.18

3 0

3 years ago

Read 2 more answers

Brian spent $36 on fruit at the grocery store. He spent a total of $40 at the store. What percentage of the total did he spend o

hammer [34]

Answer:

Just divide lol.

Step-by-step explanation:

5 0

3 years ago

Which of the following is a solution to the inequality - 2x +3>7 ?

V125BC [204]

Answer:

the solution is x<-2

Step-by-step explanation:

each time you multiply both sides with a negative number or if you divide both sides by a negative number, you need to "turn" the greater/less sign, so, transform

into

and vice versa

if you flip the sides of the equation, you also need to "turn" the greater/less sign.

6 0

3 years ago