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attashe74 [19]
3 years ago
12

you invest $300 at 4% interest, compound every year. What will your balance be after 5 years? (Remember, the formula is A = P(1

+ r)t.)
Mathematics
1 answer:
Kazeer [188]3 years ago
4 0

Answer:

$7500

Step-by-step explanation:

We have been given the formula to use but we will have to be able to recognize what each of the letters in the formula represent

A=P(1+r)t

A stands for the interest

P stands for principal

r stands for the rate

t stands for the time

So from the question we can deduce that we are to find the value A

P=$300

r=4%

t=5years

A=?

A=300(1+4)5

Using the BODMAS principle by solving the bracket first

A=300(5)5

A=300(25)

A=7500

Therefore the interest is $7500

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Step-by-step explanation:

as given,

hari bought,

amount of pure ghee he bought = 25 kg

cost of pure ghee per kg = Rs. 180

so, cost of 25 kg of pure ghee = 180 × 25 = Rs 4,500

amount of vanaspati ghee = 25 kg

cost of per kg of vanaspati ghee = Rs. 80

so, cost of 25 kg of vanaspati ghee = 80 × 25 = Rs. 2,000

he mixed both the ghee and got ( 25 + 25 ) = 50 kg of mixed ghee.

he sold them in cost = Rs. 150 per kg

so, total SP of mixed ghee = 150 × 50 = Rs. 7,500

so,

total CP = ( 4500 + 2000 ) = Rs. 6,500

total SP = Rs. 7,500

so,

Profit = SP - CP

→ 7,500 - 6,500 = Rs. 1000

Profit / gain = ( gain × 100 / CP)

→ (1000 × 100 / 6,500 )

→ 1000/65 = 15.38 ( approx.)

therefore, it's a profit of 15.38%.

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Step-by-step explanation:

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Blizzard [7]

Answer:

Correct option is (a). 60.391 and 101.879; because the test statistic is in a critical region, the test rejects the null hypothesis.

Step-by-step explanation:

A Chi-square test for population variance is used to perform this test.

The standard deviation is, 2.0 minutes.

Then the variance is, 4.0 minutes.

The hypothesis for this test is:

<em>H₀</em>: The population variance of all commute times is equal to 4.0 minutes, i.e. <em>σ² </em>= 4.

<em>Hₐ</em>: The population variance of all commute times is not equal to 4.0 minutes, i.e. <em>σ² </em>≠ 4.

The test statistic is:

\chi^{2}_{calc.}=\frac{(n-1)s^{2}}{\sigma^{2}}

The critical region of this test is defined as:

Reject <em>H₀</em> if \chi^{2}_{calc.} or \chi^{2}_{calc.}>\chi^{2}_{(1-\alpha /2), (n-1)}.

The degrees of freedom is:

n-1=81-1=80

Compute the critical from a Chi-square table.

\chi^{2}_{\alpha /2, (n-1)}=\chi^{2}_{0.05, 80}=101.879\\\chi^{2}_{(1-\alpha /2), (n-1)}=\chi^{2}_{0.95, 80}=60.391\\

The test statistic value is, \chi^{2}_{calc.}=105.8.

\chi^{2}_{calc.}=105.8 > \chi^{2}_{0.05, 80}=101.879

The null hypothesis is rejected because the test statistic is in the critical region.

Thus, the correct option is (a).

6 0
3 years ago
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