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alisha [4.7K]
3 years ago
8

How many bundles of strip flooring are needed to cover a rectangular floor area 14 ft by 13 ft 6 in in size if a bundle of strip

flooring contains 29 sq​ ft, and if 25​% extra must be allowed for side and end​ matching?
Mathematics
1 answer:
OLEGan [10]3 years ago
8 0

Answer:

Under this conditions 8.15 strips will be needed to cover the floor.

Step-by-step explanation:

In order to solve this problem we first need to calculate the area of the floor, since it's rectangular it's area is given by the product of it's length and width. In this case one of these measurements was given to us with a mixed unit, we need to convert it to a single unit in order to calculate correctly. This is done bellow:

width = 13 ft 6 in

width = 13 ft + 6 in

width = 13 ft + (6 in)/12 = 13 + 0.5 = 13.5 ft

Now we can calculate the area of the floor as shown bellow:

floor area = width*length = 14*13.5 = 189 ft²

Since each strip is 29 ft² to calculate how many strip will be used we need to divide the area of the whole floor by the area of each strip. We have:

number of strips = 189/29 = 6.52

Althought there must be 25% more strips, so we need to multiply that value by 1.25.

number of strips = 6.52*1.25 = 8.15

Under this conditions 8.15 strips will be needed to cover the floor.

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Please help!<br> Hard math.
Reil [10]

Answer:

the value of digit 3 in 156.32 =

3 x 0.1 = 0.3


The value of digit 3 in 13 =

3 x 1 = 3

3/0.3=10 times

The value of digit 3 in 13 is 10 ten times the value of digit 3 in 156.32

As long as the (digit in ones/single digit) of a number is equal to 3

The value of digit 3 of it is always 10 ten times the value of digit 3 in 156.32


5 0
3 years ago
1.2983 rounded to nearest hundredth
exis [7]

Answer:

1.3000

Step-by-step explanation:

The 9 makes the 2 move to 3. Hope this helps.

3 0
3 years ago
A highway department executive claims that the number of fatal accidents which occur in her state does not vary from month to mo
den301095 [7]

This question is incomplete, the complete question;

A highway department executive claims that the number of fatal accidents which occur in her state does not vary from month to month. The results of a study of 193 fatal accidents were recorded. Is there enough evidence to reject the highway department executive's claim about the distribution of fatal accidents between each month? Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Fatal Accidents 11 13 22 11 15 29 15 9 13 15 22 18

Step 10 of 10 : State the conclusion of the hypothesis test at the 0.1 level of significance.

Answer: H₀ : The distribution of fatal accidents between each month is same

Step-by-step explanation:

H₀ : The distribution of fatal accidents between each month is same

H₁ : The distribution of fatal accidents between each month is different

Let the los be alpha 0.10

given data ;

       Observed    Expected  

Mon   Freq (Oi)   Freq Ei       (Oi-Ei)^2 /Ei

Jan      11    16.0833        1.606649

Feb      13    16.0833        0.591105

Mar      22    16.0833        2.176598

Apr       11    16.0833        1.606649

May     15    16.0833        0.072971

Jun     29    16.0833        10.373489

Jul     15    16.0833        0.072971

Aug       9    16.0833        3.119603

Sep      13    16.0833        0.591105

Oct      15    16.0833        0.072971

Nov      22    16.0833        2.176598

Dec      18    16.0833       0.228411

Total:   193     93                22.689119

Expected Freq ⇒ 193 / 12 = 16.08333

Test Statistic, x² : 22.6891

Num Categories: 12

Degrees of freedom: 12 - 1 = 11

Critical value X² : 24.725

P-Value: 0.0195

since Chi-square value < Chi-square critical value

and P-value > alpha 0.01 so we accept H₀

Therefore we conclude that the distribution of fatal accidents between each month is same

7 0
3 years ago
The position of a particle on the x-axis at time t, t &gt; 0, is s(t) = ln(t) with t measured in seconds and s(t) measured in fe
Mrrafil [7]

Answer:

C.<em> </em><em>The quotient of the natural logarithm of 2 and e.</em>

Step-by-step explanation:

The position of a particle on the x-axis at time t, t > 0, is given by

s(t) = \ln(t)

with t is in seconds and s(t) is in feet.

The rate of change is,

=\dfrac{f(b)-f(a)}{b-a}

So rate of change of position or average velocity, for e ≤ t ≤ 2e will be,

=\dfrac{\ln2e-\ln e}{2e-e}

=\dfrac{\ln2e-\ln e}{e}

=\dfrac{\ln\frac{2e}{e}}{e}

=\dfrac{\ln2}{e}

Therefore, option C is the correct answer.

8 0
3 years ago
Read 2 more answers
PLS HELP I WILL GIVE BRANLIEST
Phantasy [73]

I did the test ma'am

5 0
3 years ago
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