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Katarina [22]
2 years ago
5

Brian invests £4900 into his bank account.

Mathematics
2 answers:
True [87]2 years ago
6 0

Answer:

  • £5200.68

Step-by-step explanation:

<u>Given:</u>

  • Investment P = £4900
  • Interest rate r = 1.5% or r = 0.015
  • Time t = 4 years
  • Number of compounds per year n = 1

<u>Find the future amount:</u>

  • F = P*(1+r/n)^{nt}\\
  • F = 4900*(1+0.015/1)^{1*4}=4900*1.015^4 = 5200.68
Reika [66]2 years ago
5 0

Answer:

£ 5200.37

Step-by-step explanation:

You can use the below formula to find the total amount of money which he will receive after n years when he invests in compound interest.

A = P [ 1 +  \frac{r}{100}  ] ^{n}

Here,

A ⇒ Amount

P ⇒ Money invested ⇒ £4900

r ⇒ Percentage ⇒ 1.5%

n ⇒ No. of years ⇒ 4 years

Let us solve now.

A = P [ 1 +  \frac{r}{100}  ] ^{n}

A =4900[1+\frac{1.5}{100} ]^{4}

A =4900[1.015]^{4}

A=4900[1.0613]

A=5200.37

Hope this helps you :-)

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A random sample of 60 mathematics majors spent an average of $200.00 for textbooks for a term, with a standard deviation of $22.
tangare [24]

Answer:

20 +/- $6.74

= ( $13.26, $26.74)

The 90% confidence interval for the difference in average amounts spent on textbooks (math majors - English majors) is ( $13.26, $26.74)

Step-by-step explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

x1-x2 +/- margin of error

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Given that;

Mean x1 = $200

x2 = $180

Standard deviation r1 = $22.50

r2 = $18.30

Number of samples n1 = 60

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Confidence interval = 90%

z(at 90% confidence) = 1.645

Substituting the values we have;

$200-$180 +/-1.645(√(22.5^2/60 +18.3^2/40)

$20 +/- 6.744449847374

$20 +/- $6.74

= ( $13.26, $26.74)

The 90% confidence interval for the difference in average amounts spent on textbooks (math majors - English majors) is ( $13.26, $26.74)

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3 years ago
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