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2 years ago

A long-distance runner finished a race in 2.75 hours. How is the runner's time expressed

Mathematics

1 answer:

MissTica2 years ago

8 0

Yes,

because 2.75 = 11/4

In mixed number it will be 2 3/4

4x2=8+3=11/4

Hopefully that makes sen to u

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If I have 2 quarters 12 dimes and 23 pennies how many more pennies will I need in order to have a toldal of $2.00

umka2103 [35]

Answer:

You will need 7 more pennies.

Step-by-step explanation:

4 0

3 years ago

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Misha deposited $3,725 into a savings account 22 years ago. The account has an interest rate of 2.2% and the balance is currentl

Annette [7]

The compounding equation is $A=P(1+ \frac{r}{n}) ^{nt}$ . So

$6012.37 = 3725(1+ \frac{0.022}{n}) ^{22n}$     Divide by 3725

$1.6141= (1+ \frac{0.022}{n}) ^{22n}$

Daily    = $(1+ \frac{0.022}{365}) ^{22*365}$
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Monthly = $(1+ \frac{0.022}{12}) ^{22*12}$
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              = 1.6218
Quarter = $(1+ \frac{0.022}{4}) ^{22*4}$
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Yearly = $(1+ \frac{0.022}{1}) ^{22*1}$
              = $(1.022) ^{22}$
              = 1.6141

The answer is yearly.

5 0

4 years ago

Latex- free gloves can be purchased in a box of 100 for $44.95 or a box of 50 for $23.50. Which one would be the best deal

malfutka [58]

The box of 100 would be the better deal, considering buying 2 boxes of 50 would turn out to be $47.00

3 0

4 years ago

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Round to the nearest hundredth : 3.2649

Bas_tet [7]

Answer:

3.26

Step-by-step explanation:

Rounding to the nearest hundredth means round to the nearest two decimal places.

To do this, all we need to look at is the 3rd decimal place.

Since 4 is not 5 or greater, that means we must round down, no matter how close it is to 5.

Therefore, the answer should just be 3.26

3 0

4 years ago

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The owner of a football team claims that the average attendance at games is over 523, and he is therefore justified in moving th

USPshnik [31]

Answer:

C) There is not sufficient evidence to support the claim that the mean attendance is greater than 523.

Step-by-step explanation:

Let μ be the the average attendance at games of the football team

The claim: the average attendance at games is over 523

Null and alternative hypotheses are:

$H_{0}$ : μ=523

$H_{a}$ : μ>523

The conclusion is failure to reject the null hypothesis.

This means that test statistic is lower than critical value. Therefore it is not significant, there is no significant evidence to accept the alternative hypothesis.

That is no significant evidence that the average attendance at games of the football team is greater than 523.

7 0

3 years ago