The total amount in the account when turning 60 years is A = $ 11,13,706.08
Given data ,
A savings account with an interest rate of 6% per year and you aren't adding any additional funds in the future
Now , you make $80,000 within the year you turn 60
So , the number of years = 60 - 16 = 44 years
And , from the compound interest , we get
A = P ( 1 + r/n )ⁿᵇ
On simplifying , we get
Where A is the final amount, P is the initial amount (which is 0 in this case), r is the annual interest rate (6% or 0.06), n is the number of times the interest is compounded per year (let's assume it is compounded monthly, so n=12), t is the time in years (44 years from age 16 to age 60).
A = 80,000 ( 1 + 0.06/12 )¹²ˣ⁴⁴
A = $ 11,13,706.08
Hence , the amount in account is A = $ 11,13,706.08
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What is the total cost of 20 books at R25 each?
Answer:
R500
Step-by-step explanation:
20 books x R25 each = R500.
How many turning points are in the graph of the polynomial function?
4 turning points
5 turning points
6 turning points
7 turning points
You roll a 6-sided number cube and toss a coin. Let event A = Toss a heads.
What outcomes are in event A?
What outcomes are in event AC?
1. Event A includes the outcomes of H and T,
2. while event AC includes all the possible outcomes of rolling a number cube, which are 1, 2, 3, 4, 5, and 6.
1. Event A is defined as tossing a heads on a coin, regardless of the outcome of rolling a number cube. Therefore, the outcomes in event A are H (heads) and T (tails), since either of these outcomes could occur when rolling a number cube and tossing a coin.
2. Event AC is the complement of event A, i.e., it is the set of outcomes that are not in event A. Since event A contains H and T, the outcomes in event AC are the remaining outcomes that are not in event A, which are all the possible outcomes when rolling a number cube: 1, 2, 3, 4, 5, and 6.
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Out of 1000 students who appeared in an examination,60% passed the examination.60% of the failing students failed in mathematics and 50% of the failing students failed in English.If the students failed in English and Mathematics only, find the number of students who failed in both subjects.
The value of number of students who failed in both mathematics and English is 40.
Since, Given that;
60% of the 1000 students passed the examination,
Hence, we can calculate the number of students who passed the exam as follows:
60/100 x 1000 = 600
So, 600 students passed the examination.
Now, let's find the number of students who failed the examination.
Since 60% of the students passed, the remaining 40% must have failed. Therefore, the number of students who failed the examination is:
40/100 x 1000 = 400
Of the 400 failing students, we know that 60% failed in mathematics.
So, the number of students who failed in mathematics is:
60/100 x 400 = 240
Similarly, we know that 50% of the failing students failed in English.
So, the number of students who failed in English is:
50/100 x 400 = 200
Now, we need to find the number of students who failed in both subjects.
We can use the formula:
Total = A + B - Both
Where A is the number of students who failed in mathematics, B is the number of students who failed in English, and Both is the number of students who failed in both subjects.
Substituting the values we have, we get:
400 = 240 + 200 - Both
Solving for Both, we get:
Both = 240 + 200 - 400
Both = 40
Therefore, the number of students who failed in both mathematics and English is 40.
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Teena uses 1/4 cup of oil for a cake. How many cakes can she make if she has 6 cups of oil?
Answer:
24 cakes.
Step-by-step explanation:
6 cups of oil divided by 1/4 cup oil per cake = 24 cakes
6/(1/4) = 24
or 6/(0.25) = 24
She can make 24 cakes with 6 cups of oil.
Instructions: Find the missing probability.
P(B)=1/2P(A|B)=11/25P(AandB)=
Please help. Is the answer even there?
The critical values t₀ for a two-sample t-test is ± 2.0.6
To find the critical values t₀ for a two-sample t-test to test the claim that the population means are equal (i.e., µ₁ = µ₂), we need to use the following formula:
t₀ = ± t_(α/2, df)
where t_(α/2, df) is the critical t-value with α/2 area in the right tail and df degrees of freedom.
The degrees of freedom are calculated as:
df = (s₁²/n₁ + s₂²/n₂)² / [(s₁²/n₁)²/(n₁-1) + (s₂²/n₂)²/(n₂-1)]
n₁ = 14, n₂ = 12, X₁ = 6,X₂ = 7, s₁ = 2.5 and s₂ = 2.8
α = 0.05 (two-tailed)
First, we need to calculate the degrees of freedom:
df = (s₁²/n₁ + s₂²/n₂)² / [(s₁²/n₁)²/(n₁-1) + (s₂²/n₂)²/(n₂-1)]
= (2.5²/14 + 2.8²/12)² / [(2.5²/14)²/13 + (2.8²/12)²/11]
= 24.27
Since this is a two-tailed test with α = 0.05, we need to find the t-value with an area of 0.025 in each tail and df = 24.27.
From a t-distribution table, we find:
t_(0.025, 24.27) = 2.0639 (rounded to four decimal places)
Finally, we can calculate the critical values t₀:
t₀ = ± t_(α/2, df) = ± 2.0639
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