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3 years ago

What is the interest on RS. 1000 at 10 p.c.p.a for 1 year.

Mathematics

2 answers:

SOVA2 [1]3 years ago

5 0

Answer:

Interest is Rs.10

Step-by-step explanation:

Interest of Rs.100 on 10 p.c.p.a is 10% of 100

i.e., 100*10/100 = 10

Thus we have 10 as the interest

Delicious77 [7]3 years ago

3 0

Answer:

Interest wil be 10% of 100 which is 10

So,

interest will be Rs.10

and total amount to be payed will be 100+10 which is Rs.110

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The temperature of lake water over a period of time is shown in the graph below.

castortr0y [4]

Answer:

Answers are 3 AM ; 10 , 54

Step-by-step explanation:

We are given a graph.

Recording of temperature starts from 9 PM.

So from Graph,

Temperature ate 9 PM = 59°F

Temperature ate 11 PM = 54°F

Temperature ate 1 AM = 51°F

Temperature ate 3 AM = 50°F

Temperature ate 5 AM = 51°F

Temperature ate 7 AM = 54°F

Temperature ate 9 AM = 59°F

So, At 3 AM the water is at its lowest temperature.

and

After 10 hours passed, the temperature of the lake is 54°F.

Therefore, Answers are 3 AM ; 10 , 54

7 0

3 years ago

Read 2 more answers

A random sample of 150 men found that 88 of the men excercise regularly, while a random sample 200 women found that 130 of the w

melamori03 [73]

Answer:

The hypothesis is:

H₀: $p_{X}-p_{Y}=0$ .

Hₐ: $p_{X}-p_{Y}$ .

Step-by-step explanation:

Let X = number of men who exercise regularly and Y = number of women who exercise regularly.

The information provided is:

$n_{X}=150\\X=88\\n_{Y}=200\\Y=130$

Compute the sample proportion of men and women who exercise regularly as follows:

$\hat p_{X}=\frac{X}{n_{X}}=\frac{88}{150}=0.587$

$\hat p_{Y}=\frac{Y}{n_{Y}}=\frac{130}{200}=0.65$

The random variable X follows a Binomial distribution with parameters n = 150 and $\hat p_{X}=0.587$ .

The random variable Y also follows a Binomial distribution with parameters n = 200 and $\hat p_{Y}=0.65$ .

According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.

The mean of this sampling distribution of sample proportion is:

$\hat p=p$

The standard deviation of this sampling distribution of sample proportion is:

$\sigma_{\hat p}=\sqrt{\frac{\hat p(1-\hat p)}{n}}$

So, the sampling distribution of the proportion of men and women who exercise regularly follows a Normal distribution.

A two proportion z-test cab be performed to determine whether the proportion of women is more than men who exercise regularly.

The hypothesis for this test cab be defined as:

H₀: The proportion of women is same as men who exercise regularly, i.e. $p_{X}-p_{Y}=0$ .

Hₐ: The proportion of women is more than men who exercise regularly, i.e. $p_{X}-p_{Y}$ .

6 0

3 years ago

Read 2 more answers

Nikki spent 2/5 of her birthday money on new shoes and 1/4 of her birthday money on a new headband.

quester [9]

Answer:

/20

Step-by-step explanation

7 0

2 years ago

Which expression is NOT equivalent to 24·−458?

maks197457 [2]

Answer:

Step-by-step explanation:

7 0

3 years ago

A radioactive substance decays exponentially. A scientist begins with 140 milligrams of a radioactive substance. After 25 hours,

zmey [24]

Answer:

Therefore 53.05 mg will remain of the given radioactive substance after 35 hours.

Step-by-step explanation:

Radioactive Decay:

$\frac{dN}{dt}\propto N$

$\Rightarrow \frac{dN}{dt}=\lambda N$

$\Rightarrow \frac{dN}{N}=\lambda dt$

Integrating both sides

$\int \frac{dN}{N}=\int\lambda dt$

$\Rightarrow ln |N|= \lambda t+c_1$

$\Rightarrow N= e^{\lambda t+c_1}$

$\Rightarrow N= e^{\lambda t}.e^{c_1}$

$\Rightarrow N=c e^{\lambda t}$ [ $e^{c_1}=c$ ]

When t=0, $N=N_0$ = initial amount

$N_0=c e^{\lambda .0}$

$\Rightarrow N_0=c$

Therefore the decay equation is

$N=N_0e^{\lambda t}$

Given that, $N_0$ = Initial amount of the radioactive substance= 140 mg

After 25 hours, 70 mg of substance remains.

N= 70 mg, t=25 hours

$N=N_0e^{\lambda t}$

$\Rightarrow 70 =140e^{\lambda \times 25}$

$\Rightarrow e^{\lambda \times 25}=\frac{70}{140}$

$\Rightarrow ln|e^{\lambda \times 25}|=ln|\frac{1}{2}|$

$\Rightarrow \lambda \times 25}=-ln|2|$ [ $ln |\frac12|=ln 1-ln 2=0-ln 2=-ln 2$ ]

$\Rightarrow \lambda =-\frac{ln|2|}{25}$

The decay equation becomes

$N=N_0e^{-\frac{ln|2|}{25}t}$

Now putting t= 35

$N=140 e^{-\frac{ln|2|}{25}.35}$

= 53.05 mg

Therefore 53.05 mg will remain of the given radioactive substance after 35 hours.

3 0

3 years ago

What is the interest on RS. 1000 at 10 p.c.p.a for 1 year.​

What is the interest on RS. 1000 at 10 p.c.p.a for 1 year.