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3 years ago

1. How many cups of sugar will it take if you had to bake 10 cakes and each cake requires 4 cups of sugar?

Mathematics

2 answers:

eimsori [14]3 years ago

7 0

Answer:

9 bags of suger

Step-by-step explanation:

goblinko [34]3 years ago

4 0

Answer:

For the first one you need 40 cups of sugar and for the second one you need 80.

Step-by-step explanation:

Simple multiplication.

10x4 = 40

10x8 = 80

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Surface area help plz

lana [24]

Answer:

486 mm

Step-by-step explanation:

solve each side on it's own first

First 2 rectangles: 288 mm all together

2 rectangle: 144 mm

Triangles: 54 mm

Then add rectangles and triangles:

288 mm + 144 mm + 54 mm = 486 mm

8 0

3 years ago

The frog jumped 7/8 of a foot on the first jump and 2/3 of a foot on the second

Marina86 [1]

He jumped 1.542 feet
7/8 + 2/3

5 0

3 years ago

7.4 Practice

pickupchik [31]

The simulation of the medicine and the bowler hat are illustrations of probability

The probability that the medicine is effective on at least two is 0.767
The probability that the medicine is effective on none is 0
The probability that the bowler hits a headpin 4 out of 5 times is 0.3281

<h3>The probability that the medicine is effective on at least two</h3>

From the question,

Numbers 1 to 7 represents the medicine being effective
0, 8 and 9 represents the medicine not being effective

From the simulation, 23 of the 30 randomly generated numbers show that the medicine is effective on at least two

So, the probability is:

p = 23/30

p = 0.767

Hence, the probability that the medicine is effective on at least two is 0.767

<h3>The probability that the medicine is effective on none</h3>

From the simulation, 0 of the 30 randomly generated numbers show that the medicine is effective on none

So, the probability is:

p = 0/30

p = 0

Hence, the probability that the medicine is effective on none is 0

<h3>The probability a bowler hits a headpin</h3>

The probability of hitting a headpin is:

p = 90%

The probability a bowler hits a headpin 4 out of 5 times is:

P(x) = nCx * p^x * (1 - p)^(n - x)

So, we have:

P(4) = 5C4 * (90%)^4 * (1 - 90%)^1

P(4) = 0.3281

Hence, the probability that the bowler hits a headpin 4 out of 5 times is 0.3281

Read more about probabilities at:

brainly.com/question/25870256

8 0

2 years ago

Problem Page Donna had 2/7 acres of land. She planted 4/5 of it with corn. How many acres did she plant with corn?

Dafna11 [192]

Acres of planted corn = (2/7) * (4/5)

= (2/7) * .8 = 1.6 / 7 acres = 0.2285714286 acres

5 0

4 years ago

How many nonzero terms of the Maclaurin series for ln(1 x) do you need to use to estimate ln(1.4) to within 0.001?

Vilka [71]

Answer:

The estimate of In(1.4) is the first five non-zero terms.

Step-by-step explanation:

From the given information:

We are to find the estimate of In(1 . 4) within 0.001 by applying the function of the Maclaurin series for f(x) = In (1 + x)

So, by the application of Maclurin Series which can be expressed as:

$f(x) = f(0) + \dfrac{xf'(0)}{1!}+ \dfrac{x^2 f"(0)}{2!}+ \dfrac{x^3f'(0)}{3!}+... \ \ \ \ \ --- (1)$

Let examine f(x) = In(1+x), then find its derivatives;

f(x) = In(1+x)

$f'(x) = \dfrac{1}{1+x}$

f'(0) $= \dfrac{1}{1+0}=1$

f ' ' (x) $= \dfrac{1}{(1+x)^2}$

f ' ' (x) $= \dfrac{1}{(1+0)^2}=-1$

f ' ' '(x) $= \dfrac{2}{(1+x)^3}$

f ' ' '(x) $= \dfrac{2}{(1+0)^3} = 2$

f ' ' ' '(x) $= \dfrac{6}{(1+x)^4}$

f ' ' ' '(x) $= \dfrac{6}{(1+0)^4}=-6$

f ' ' ' ' ' (x) $= \dfrac{24}{(1+x)^5} = 24$

f ' ' ' ' ' (x) $= \dfrac{24}{(1+0)^5} = 24$

Now, the next process is to substitute the above values back into equation (1)

$f(x) = f(0) + \dfrac{xf'(0)}{1!}+ \dfrac{x^2f' \ '(0)}{2!}+\dfrac{x^3f \ '\ '\ '(0)}{3!}+\dfrac{x^4f '\ '\ ' \ ' \(0)}{4!}+\dfrac{x^5f' \ ' \ ' \ ' \ '0)}{5!}+ ...$

$In(1+x) = o + \dfrac{x(1)}{1!}+ \dfrac{x^2(-1)}{2!}+ \dfrac{x^3(2)}{3!}+ \dfrac{x^4(-6)}{4!}+ \dfrac{x^5(24)}{5!}+ ...$

$In (1+x) = x - \dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}- \dfrac{x^6}{6}+...$

To estimate the value of In(1.4), let's replace x with 0.4

$In (1+x) = x - \dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}- \dfrac{x^6}{6}+...$

$In (1+0.4) = 0.4 - \dfrac{0.4^2}{2}+\dfrac{0.4^3}{3}-\dfrac{0.4^4}{4}+\dfrac{0.4^5}{5}- \dfrac{0.4^6}{6}+...$

Therefore, from the above calculations, we will realize that the value of $\dfrac{0.4^5}{5}= 0.002048$ as well as $\dfrac{0.4^6}{6}= 0.00068267$ which are less than 0.001

Hence, the estimate of In(1.4) to the term is $\dfrac{0.4^5}{5}$ is said to be enough to justify our claim.

∴

The estimate of In(1.4) is the first five non-zero terms.

8 0

3 years ago