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SpyIntel [72]
2 years ago
12

A carpet measures 7 meters 3 decimeters in length and 4 meters 2 decimeters in width. What is its area in square meters?

Mathematics
2 answers:
Varvara68 [4.7K]2 years ago
8 0

Answer:

C

Step-by-step explanation:

change decimetres into metres

length - 7.3m

Width - 4.2m

to find an area, do length times width

7.3 x 4.2 = 30.66 square metres

I hope this helped :)

Semmy [17]2 years ago
6 0

Answer:

C. 30.66 square meters

Step-by-step explanation:

The equation for area is A = L × W

L = length

W = width

Because a demimeter is 1 tenth of a meter your new measurements are 7.3 and 4.2.

multiply the measurements to find the area of 30.66² meters

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Draw a number line and mark all described points. 3x<8
Leya [2.2K]

3x8 is 24 hopefully this is helpful

7 0
3 years ago
Read 2 more answers
The article "Students Increasingly Turn to Credit Cards" (San Luis Obispo Tribune, July 21, 2006) reported that 37% of college f
Sloan [31]

Answer:

Step-by-step explanation:

Hello!

There are two variables of interest:

X₁: number of college freshmen that carry a credit card balance.

n₁= 1000

p'₁= 0.37

X₂: number of college seniors that carry a credit card balance.

n₂= 1000

p'₂= 0.48

a. You need to construct a 90% CI for the proportion of freshmen  who carry a credit card balance.

The formula for the interval is:

p'₁±Z_{1-\alpha /2}*\sqrt{\frac{p'_1(1-p'_1)}{n_1} }

Z_{1-\alpha /2}= Z_{0.95}= 1.648

0.37±1.648*\sqrt{\frac{0.37*0.63}{1000} }

0.37±1.648*0.015

[0.35;0.39]

With a confidence level of 90%, you'd expect that the interval [0.35;0.39] contains the proportion of college freshmen students that carry a credit card balance.

b. In this item, you have to estimate the proportion of senior students that carry a credit card balance. Since we work with the standard normal approximation and the same confidence level, the Z value is the same: 1.648

The formula for this interval is

p'₂±Z_{1-\alpha /2}*\sqrt{\frac{p'_2(1-p'_2)}{n_2} }

0.48±1.648* \sqrt{\frac{0.48*0.52}{1000} }

0.48±1.648*0.016

[0.45;0.51]

With a confidence level of 90%, you'd expect that the interval [0.45;0.51] contains the proportion of college seniors that carry a credit card balance.

c. The difference between the width two 90% confidence intervals is given by the standard deviation of each sample.

Freshmen: \sqrt{\frac{p'_1(1-p'_1)}{n_1} } = \sqrt{\frac{0.37*0.63}{1000} } = 0.01527 = 0.015

Seniors: \sqrt{\frac{p'_2(1-p'_2)}{n_2} } = \sqrt{\frac{0.48*0.52}{1000} }= 0.01579 = 0.016

The interval corresponding to the senior students has a greater standard deviation than the interval corresponding to the freshmen students, that is why the amplitude of its interval is greater.

8 0
3 years ago
On Mars, an object weighs 60% of its weight on Earth. Shane weighs 125 pounds on Earth.
Margarita [4]

A and D I think or it may just be A

6 0
2 years ago
Kishore bought a basket containing 30 orangs for rs 310. If he sells the orangs at rs 12 each , what Would be it's profit or los
iris [78.8K]

Given : Kishore bought a basket containing 30 oranges for rs 310. If he sells the oranges at rs 12 each , what Would be it's profit or loss?

Solution :

Cost price of oranges = rs.310

As we know that each oranges cost rs.12

So, selling price of oranges is 30 × 12 = rs.360

Now, <u>finding profit or loss</u>

Profit = Selling Price - Cost Price

Profit = rs.360 - rs.310

Profit = rs.50

  • So, the profit is ₹50
7 0
2 years ago
You estimate that there are 44 marbles in a jar. The actual amount is 73 marbles. Find the percent error.
Svetlanka [38]

Answer:

Percentage Error in counting the marbles in the jar is 39.72%

Step-by-step explanation:

Estimated marbles in the jar = 44

Actual marbles in the jar  = 73

⇒Error in  Marbles  =  Number of actual Marbles - Estimated marbles

                                 =   73  - 44 = 29

So, the error in counting marbles in the jar is 29.

Now, \textrm{Percentage Error}  = \frac{\textrm{Error in the Data}}{\textrm{Actual Data}}  \times 100

= \frac{29}{73}  \times 100  = 39.72

or, Percentage Error  =  39.72%

Hence, the Percentage Error in counting the marbles in the jar is 39.72%

5 0
2 years ago
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