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Snowcat [4.5K]
2 years ago
10

During a certain week, a post office sold Rs.280 worth of 14-paisas stamps. How many of these stamps did they sell?

Mathematics
1 answer:
Novosadov [1.4K]2 years ago
6 0
So basically ...

You convert the rupees in paisas. One rupee is equal to one hundred paisas, so ...

280 × 100 = 28,000

And then we divide,

28,000 ÷ 14 = 2000

The post office sold 2000 stamps!

Hope this helped! :)
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A stadium asked for $760,000 for naming rights. During a bidding war for the stadium naming rights, one company bid 7% more than
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Answer:

\$\ 829464

Step-by-step explanation:

Bid of first company:

Bid is 7\% more

7\% of 760000=\frac{7}{100}\times760000=53200

Bid of first =760000+53200=813200

Bid of second company:

Bid of second is 2\% greater than that of first

2\% of first bid =\frac{2}{100}\times 813200=16264

Hence Bid of second company =813200+16264=829464

8 0
2 years ago
NEED HELP FAST WILL GIVE BRAINIEST
vladimir1956 [14]

Answer:

x^2 + 8x - 65 = 0.

Step-by-step explanation:

In order to solve the side of the enlarged area of 81 square inches, we use the equation in solving the area of a square. Area = side^2.

the increase in side will be x, so (4+x)^2 = 81, x^2 + 4x + 4x + 16 = 81

x^2 + 8x + 16 - 81 = 0

x^2 + 8x - 65 = 0.

  Hoped this helped mark Brainliest!

5 0
2 years ago
Read 2 more answers
Body armor provides critical protection for law enforcement personnel, but it does affect balance and mobility. The article "Imp
yKpoI14uk [10]

Answer:

No, there is not enough evidence to support the claim that true average task time with armor is less than 2 seconds.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that true average task time with armor is less than 2 seconds.

Then, the null and alternative hypothesis are:

H_0: \mu=2\\\\H_a:\mu< 2

The significance level is 0.01.

The sample has a size n=52.

The sample mean is M=1.95.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=0.2.

The estimated standard error of the mean is computed using the formula:

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{0.2}{\sqrt{52}}=0.028

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{1.95-2}{0.028}=\dfrac{-0.05}{0.028}=-1.803

The degrees of freedom for this sample size are:

df=n-1=52-1=51

This test is a left-tailed test, with 51 degrees of freedom and t=-1.803, so the P-value for this test is calculated as (using a t-table):

P-value=P(t

As the P-value (0.039) is bigger than the significance level (0.01), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that true average task time with armor is less than 2 seconds.

3 0
2 years ago
The heights of 200 adults were recorded and divided into two categories. Which two-way frequency table correctly shows the margi
Temka [501]

Answer: c

Step-by-step explanation:

C

7 0
2 years ago
A poll conducted a week before the school election to the student council showed that Janice would win with 63% of the vote. The
k0ka [10]

OPTIONS:

No, because she could receive as low as 14% of the vote.

No, because she could receive as low as 49% of the vote.

Yes, because she could receive as much as 77% of the vote.

Yes, because the poll stated that she will win with 63% of the vote.

Answer:

No, because she could receive as low as 49% of the vote.

Step-by-step explanation:

Given that:

Poll suggestion = 63%

Margin of Error (E) = 14%

This means that; the range or interval in which the vote obtained could fall would be ;

Poll suggestion ± margin of Error

63% ± 14%

Lower bound attainable = (63% - 14%) = 49%

Upper bound attainable = (63% + 14%) = 77%

Since, winning actually requires obtaining atleast half of the votes (that is 50%) ; and suggested poll suggested votes could fall as low as 49% ; then Janice can't be so confident of victory.

8 0
2 years ago
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