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2 months ago

Which of these examples is an error of addition by one?

Mathematics

2 answers:

tester [12.3K]2 months ago

8 0

My response will be B because 38,273+51,425=89698.

Inessa [12.5K]2 months ago

4 0

The answer is B. 38,273 + 51,425 equals 89,798. Step-by-step explanation: Let's verify each sum by adding up their respective places. A: It's accurate. B: Here, when adding the digits in the hundredth place, we get 6 instead of 7. Therefore, this is an example of an addition error by one. It is indeed accurate. C: It's accurate. D: It's also accurate.

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Q13. A stone is dropped from a height of 5 km. The distance it falls through varies directly with the

Leona [12618]

I’m not really certain, but I think C could be the answer; my apologies if that's inaccurate.

8 0

3 months ago

Read 2 more answers

−7x−50≤−1 AND−6x+70>−2

PIT_PIT [12445]

Answer:

$-7\geq x$ and $[-7,12)$ expressed in interval notation.

Step-by-step explanation:

A compound inequality $-7x-50\leq -1\text{ and }-6x+70>-2$ has been provided. Our task is to determine the solution for this inequality.

Initially, we will address each inequality independently, followed by merging the findings by combining the overlapping intervals.

$-7x-50\leq -1$

$-7x-50+50\leq -1+50$

$-7x\leq 49$

By dividing with a negative number, it is necessary to reverse the inequality sign:

$\frac{-7x}{-7}\geq \frac{49}{-7}$

$x\geq -7$

$-6x+70>-2$

$-6x+70-70>-2-70$

$-6x>-72$

Again, dividing by a negative requires flipping the inequality sign:

$\frac{-6x}{-6}$

$x$

In combining both intervals, we will arrive at:

$-7\geq x$

Thus, the solution for the inequality provided is $-7\geq x$ and $[-7,12)$ in interval notation.

7 0

3 months ago

Let c1(t) = eti + (sin(t))j + t3k and c2(t) = e−ti + (cos(t))j − 6t3k. Find the stated derivatives in two different ways to veri

Zina [12379]

Answer:

$i ( e^{t} - e^{-t})+ j (cost-sin t)+ k (-15t^{2})$

$\frac{d}{dx}(e^x) = e^x$

Step-by-step explanation:

Step 1:-

We have c1(t) = e^ t i + (sin(t))j + t³k

and c2(t) = e^−t i + (cos(t))j − 6t³k.

By adding c1(t) and c2(t):

c1(t)+c2(t) = e^ t i + (sin(t))j + t³k + e^−t i + (cos(t))j − 6t³k

Now, employing the derivative formula:

$\frac{d}{dx}(e^x) = e^x$

$\frac{d}{dx}(sinx) = cosx\\\frac{d}{dx}(cosx) = -sinx$

Next, differentiate with respect to 't'

$\frac{d}{dt}c_{1}+ c_{2} } = e^ t i +cost j +3t^2 k - e^-t i - sintj -18t^2 k$

By factoring out i, j, and k terms, we arrive at:

$\frac{d}{dt}(C_{1} +C_{2} ) = i ( e^{t} - e^{-t})+ j (cost-sin t)+ k (-15t^{2})$

7 0

2 months ago

At a small coffee shop, the distribution of the number of seconds it takes for a cashier to process an order is approximately no

Leona [12618]

Answer:

The correct answer is;

A. 0.17

Step-by-step explanation:

Here are the provided details;

The average time taken for a cashier to handle an order, μ = 276 seconds

The deviation from the average, σ = 38 seconds

The z-score for an order processing time of x = 240 seconds can be calculated as follows;

$Z=\dfrac{x-\mu }{\sigma }$

Thus;

$Z=\dfrac{240-276 }{38 } \approx -0.9474$

The resulting probability

P(z = -0.9474) = 0.17361

Hence, the estimated proportion of orders processed in under 240 seconds is roughly 0.17361 or 0.17 when rounded to two decimal places.

8 0

3 months ago

If the volume of a cube is increased by a factor of 10, by what factor does the surface area per unit volume change

tester [12383]

Answer:

(1/10)∛100 ≈ 0.4642

Step-by-step explanation:

For a cubic shape with volume V, the edge length is defined as...

s = ∛V

and the surface area is given by...

A = 6s² = 6V^(2/3)

The area-to-volume ratio is therefore...

r1 = A/V = 6V^(2/3)/V = 6V^(-1/3)

When the volume V is increased by a factor of 10, the new area-to-volume ratio becomes...

r2 = 6(10V)^(-1/3)

Consequently, the change factor for the ratio is...

r2/r1 = (6(10V)^(-1/3))/(6V^(-1/3)) = 10^(-1/3) = (1/10)∛100

The change in surface area per unit volume results in a factor of (∛100)/10.

___

Example

For a cube with side length 2, the corresponding volume equals 2³ = 8, with a surface area of 6·2² = 24. The resulting area-to-volume ratio is 24/8 = 3.

The area-to-volume ratio changed by a factor of (0.3∛100)/3 = (∛100)/10, as previously noted.

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8 0

2 months ago