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andrew11 [14]
3 years ago
14

irene thinks she has space for 45 inch wide book case. it turns out she only has space for 40 inch wide book case what is the pe

rcent error in irenes measurement
Mathematics
1 answer:
TiliK225 [7]3 years ago
5 0

Answer:

Apply the percent error formula:

(Experimental-Theoretical)/Theoretical

(45-40)/40

The percent error of her measurement was 12.5%

Step-by-step explanation:

Hope this helped

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Find the product of 4/6-2a x 3-a/2<br>a.1<br>b.2<br>c.3<br>d.4
kherson [118]

Answer:

c

Step-by-step explanation:

3 0
3 years ago
the cost for a company to produce x television in 1 year is $150x + $250,000. how many televisions can the company produce in 1
vampirchik [111]

Answer:

Number of television company produce = 2,000

Step-by-step explanation:

Given:

Cost function = $150x + $250,000

Cost = $550,000

Find:

Number of television company produce

Computation:

$150x + $250,000 = $550,000

$150x = $300,000

x = 2,000

Number of television company produce = 2,000

5 0
3 years ago
If I 50% increase is followed by a 33 1\3 % decrease is it greater, less than, or the same as the original
luda_lava [24]

Answer:

Greater than.

Step-by-step explanation:

Let the original value be x.

A 50% increase in x is given by

x + 50%x

=> x + 0.5x

=> 1.5x

Followed by a 33 \frac{1}{3} % decrease in the value gives:

1.5x - 33 \frac{1}{3} %(1.5x)

=> 1.5x - 0.33(1.5x)

=> 1.5x - 0.495x

=> 1.005x

Since 1.005x is greater than the original value, x, then:

A 50% increase followed by a 33 \frac{1}{3} % decrease gives a value that is greater than the original.

5 0
3 years ago
3 Identify, if the question is
nlexa [21]

Answer:

the question is nonstatistical.

3 0
4 years ago
Suppose the integral from 2 to 8 of g of x, dx equals 12, and the integral from 6 to 8 of g of x, dx equals negative 3, find the
Artyom0805 [142]

\displaystyle\int_2^8g(x)\,\mathrm dx=12

\displaystyle\int_6^8g(x)\,\mathrm dx=-3

Use the fact that integrals are additive on their intervals. Mathematically, if c\in[a,b], then

\displaystyle\int_a^bg(x)\,\mathrm dx=\int_a^cg(x)\,\mathrm dx+\int_c^bg(x)\,\mathrm dx

So we have

\displaystyle2\int_2^6g(x)\,\mathrm dx=2\left(\int_2^8g(x)\,\mathrm dx-\int_6^8g(x)\,\mathrm dx\right)=2(12-(-3))=30

8 0
4 years ago
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