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Alex Ar [27]
2 years ago
8

Dan invests £2200 into a savings account. The bank gives 5% compound interest for the first 3 years and 4% thereafter. How much

will Dan have after 7 years to the nearest pound?
Mathematics
2 answers:
Naily [24]2 years ago
7 0
Present value = 2200
interest rates =5% for 3 years, and 4% afterwards
no. of periods = 7

Future value=2200*(1.05)^3(1.04)^(7-3)
=<span>£</span>2979.367   (please round to appropriate number of decimals)
Brilliant_brown [7]2 years ago
7 0

The <em><u>correct answer</u></em>, to the nearest pound, is:

2979

Explanation:

Compound interest can be written as an equation of the form

y=a(1+r)^t,

where <em>a</em> is the initial deposit, r is the interest rate as a decimal number, and t is the number of years.

For the first 3 years, we have <em>a</em> = 2200, r = 5% = 5/100 = 0.05; and t = 3:

y=2200(1+0.05)^3\\\\=2200(1.05)^3=2546.775

This will be the "initial deposit" in the account with 4% interest.  This gives us that

<em>a</em> = 2546.775; r = 4% = 4/100 = 0.04; and t = 7-3 = 4:

y=2546.775(1+0.04)^4\\\\=2546.775(1.04)^4=2979.367 \approx 2979

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What is the perimeter of the trapezoid with vertices Q(8, 8), R(14, 16), S(20, 16), and T(22, 8)? Round to the nearest hundredth
uysha [10]

Answer:

The perimeter of the trapezoid is 38.25\ units

Step-by-step explanation:

we know that

The perimeter of the trapezoid is the sum of its four side lengths

so

In this problem

P=QR+RS+ST+QT

the formula to calculate the distance between two points is equal to

d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}

we have

Q(8, 8), R(14, 16), S(20, 16),T(22, 8)

step 1

Find the distance QR

Q(8, 8), R(14, 16)

substitute the values in the formula

d=\sqrt{(16-8)^{2}+(14-8)^{2}}

d=\sqrt{(8)^{2}+(6)^{2}}

d=\sqrt{100}

QR=10\ units

step 2

Find the distance RS

R(14, 16), S(20, 16)

substitute the values in the formula

d=\sqrt{(16-16)^{2}+(20-14)^{2}}

d=\sqrt{(0)^{2}+(6)^{2}}

d=\sqrt{36}

RS=6\ units

step 3

Find the distance ST

S(20, 16),T(22, 8)

substitute the values in the formula

d=\sqrt{(8-16)^{2}+(22-20)^{2}}

d=\sqrt{(-8)^{2}+(2)^{2}}

d=\sqrt{68}

ST=8.25\ units

step 4

Find the distance QT

Q(8, 8),T(22, 8)

substitute the values in the formula

d=\sqrt{(8-8)^{2}+(22-8)^{2}}

d=\sqrt{(0)^{2}+(14)^{2}}

d=\sqrt{196}

QT=14\ units

step 5

Find the perimeter

P=10+6+8.25+14=38.25\ units

6 0
2 years ago
Read 2 more answers
the time taken by a student to the university has been shown to be normally distributed with mean of 16 minutes and standard dev
Naya [18.7K]

Answer:

a) 2.84% probability that he is late for his first lecture.

b) 5.112 days

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

\mu = 16, \sigma = 2.1

a. Find the probability that he is late for his first lecture.

This is the probability that he takes more than 20 minutes to walk, which is 1 subtracted by the pvalue of Z when X = 20. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{20 - 16}{2.1}

Z = 1.905

Z = 1.905 has a pvalue of 0.9716

1 - 0.9716 = 0.0284

2.84% probability that he is late for his first lecture.

b. Find the number of days per year he is likely to be late for his first lecture.

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0.0284*180 = 5.112 days

4 0
2 years ago
Given triangle JKL find m
timama [110]
I would set all 3 given measures equal to 180.
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19x +9 = 180
19x = 171
x= 9

angle l = 4 (9) + 11= 47
answer = 47
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