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Anon25 [30]
3 years ago
14

If Aizuddin borrowed RM 6.300 from a bank which offers an interest of 8%

Mathematics
1 answer:
Novosadov [1.4K]3 years ago
6 0

Answer:

(a) The formula to calculate the amount of money (A) that Aizuddin must pay the bank after n years, with the original amount of borrowed money is 6300 RM, interest of 8%, compounded annually, is described as following:

A = principal x (1 + rate)^(time in year)

A = 6300 x (1 + 8/100)^n

(b) The amount of interest charged (AC) that Aizuddin must pay after n years:

AC = A - 6300

AC = 6300 x (1 + 8/100)^n - 6300

AC = 6300 x [(1 + 8/100)^n - 1]

Hope this helps!

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You roll a pair of six-sided number cubes, numbered 1 through 6. What is the probability of rolling two odd numbers? (Enter your
lbvjy [14]

Answer:

1/4 or 25%

Step-by-step explanation:

The probability for rolling an odd number is 50% or  \frac{1}{2} because there are 3 odd numbers, (1, 3, 5) and 6 total outcomes (1, 2, 3, 4, 5, 6). \frac{3}{6} =\frac{1}{2}

The probability for rolling two odd numbers in a row is

\frac{1}{2} *\frac{1}{2} =\frac{1}{4}

So, the probability of rolling an odd number twice is 25% or  \frac{1}{4}.

8 0
3 years ago
Can y’all help me on question 7 please?
Verizon [17]
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4 0
3 years ago
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The arch beneath a bridge is​ semi-elliptical, a​ one-way roadway passes under the arch. The width of the roadway is 38 feet and
forsale [732]

Answer:

Only truck 1 can pass under the bridge.

Step-by-step explanation:

So, first of all, we must do a drawing of what the situation looks like (see attached picture).

Next, we can take the general equation of an ellipse that is centered at the origin, which is the following:

\frac{x^2}{a^2}+\frac{y^2}{b^2}

where:

a= wider side of the ellipse

b= shorter side of the ellipse

in this case:

a=\frac{38}{2}=19ft

and

b=12ft

so we can go ahead and plug this data into the ellipse formula:

\frac{x^2}{(19)^2}+\frac{y^2}{(12)^2}

and we can simplify the equation, so we get:

\frac{x^2}{361}+\frac{y^2}{144}

So, we need to know if either truk will pass under the bridge, so we will match the center of the bridge with the center of each truck and see if the height of the bridge is enough for either to pass.

in order to do this let's solve the equation for y:

\frac{y^{2}}{144}=1-\frac{x^{2}}{361}

y^{2}=144(1-\frac{x^{2}}{361})

we can add everything inside parenthesis so we get:

y^{2}=144(\frac{361-x^{2}}{361})

and take the square root on both sides, so we get:

y=\sqrt{144(\frac{361-x^{2}}{361})}

and we can simplify this so we get:

y=\frac{12}{19}\sqrt{361-x^{2}}

and now we can evaluate this equation for x=4 (half the width of the trucks) so:

y=\frac{12}{19}\sqrt{361-(8)^{2}}

y=11.73ft

this means that for the trucks to pass under the bridge they must have a maximum height of 11.73ft, therefore only truck 1 is able to pass under the bridge since truck 2 is too high.

5 0
3 years ago
A hotel rents 210 rooms at a rate of $ 60 per day. For each $ 2 increase in the rate, three fewer rooms are rented. Find the roo
lidiya [134]

Answer:

r=$14,400

The hotel should charge $120

Step-by-step explanation:

Revenue (r)= p * n

where,

p = price per item

n = number of items sold

A change in price leads to a change in number sold

A variable to measure the change in p and n needs to be introduced

Let the variable=x

Such that

p + x means a one dollar price increase

p - x means a one dollar price decrease

n + x means a one item number-sold increase

n - x means a one item number-sold decrease

for each $2 price increase (p + 2x) there are 3 fewer rooms are rented (n-3x)

know that at $60 per room, the hotel rents 210 rooms

r = (60 + 2x) * (210 - 3x)

=12,600-180x+420x-6x^2

=12,600+240x-6x^2

r=2100+40x-x^2

= -x^2 +40x+2100=0

Solve the quadratic equation

x= -b +or- √b^2-4ac / 2a

a= -1

b=40

c=2100

x= -b +or- √b^2-4ac / 2a

= -40 +or- √(40)^2 - (4)(-1)(2100) / (2)(-1)

= -40 +or- √1600-(-8400) / -2

= -40 +or- √ 1600+8400 / -2

= -40 +or- √10,000 / -2

= -40 +or- 100 / -2

x= -40+100/-2 OR -40-100/-2

=60/-2 OR -140/-2

= -30 OR 70

x=70

The quadratic equation has a maximum at x=70

p+2x

=60+2(30)

=60+60

=$120

r= (60 + 2x) * (210 - 3x)

={60+2(30)}*{(210-3(30)}

r=(60+60)*(210-90)

=120*120

=$14,400

7 0
3 years ago
Two boats A and B Left a Port at the same time on different routes . B travelled on a bearing of 150º and A travelled on the Nor
adell [148]

Answer: 67.2°

Step-by-step explanation:

To find angle C, we have to use cosine rule. Therefore,

12² = 8² + 10² - 2 × 8 × 10 × cosC

C = 82.8°

The bearing of A from C will be

= 150° - 82.8°

= 67.2°.

6 0
3 years ago
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