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3 years ago

Amy knits 1/10 of a scarf in 4/5 of an hour. What fraction of scarf can Amy knit in 1 hour

Mathematics

1 answer:

g100num [7]3 years ago

6 0

Answer:

Amy can knit $\frac{1}{8}$ of a scarf in 1 hour.

Step-by-step explanation:

This question can be solved by proportions, using a rule of three.

We have that:

1/10 of a scarf - 4/5 of an hour

x of a scarf - 1 hours.

Applying cross multiplication:

$\frac{4x}{5} = \frac{1}{10}$

$x = \frac{5}{10*4} = \frac{5}{40} = \frac{1}{8}$

Amy can knit $\frac{1}{8}$ of a scarf in 1 hour.

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If f(x) = 2x + 3 and g(x) = x2-7, find (f+g)(x)

love history [14]

f(x) = 2x + 3
g(x) = x²-7

(f+g)(x) = f(x) + g(x) = 2x + 3 + x² - 7 = x² + 2x - 4

(f+g)(x) = x² + 2x - 4

3 0

3 years ago

Snowy's Snow Cones has a special bubble gum snow cone on sale. The cone is a regular snow cone that has a spherical piece of bub

cupoosta [38]

Let us first note down all the information's that are given in the question.

Radius of the snow cone = 4 inches

Height of the snow cone = 6 inches

Diameter of the bubble gum = 0.8 inches

Then

Radius of the bubble gum = 0.4 inches

Now

Volume of the come = 1/3 * pi * (radius)2 * height

= 1/3 * 3.14 * (4) ^2 * 6

= 3.14 * 16 * 2

= 100.48 cubic inches

Again

Volume of the sphere = (4/3) * pi * (radius)^3

= (4/3) * 3.14 * (0.4)^3

= (4/3) * 3.14 * 0.064

= 0.2679 cubic inches

Volume of the cone

that can be filled with flavored ice = 100.48 - 0.27

= 100.21 cubic inches

The correct option among all the options that are given in the question is option "d".

3 0

3 years ago

Consider a population list x with μx=10 and SDx = 1. A second population list, y, with μy=10 and SDy=2, is added to the first li

Basile [38]

Answer:

The combined standard deviation is 1.58114.

Step-by-step explanation:

The formula to compute the combined standard deviations of two different data sets is:

$SD_{c} =\sqrt{\frac{n_{X}S^{2}_{X}+n_{2}S^{2}_{Y}+n_{X}(\mu_{X}-\mu_{c})^{2}+n_{Y}(\mu_{Y}-\mu_{c})^{2}}{n_{X}+n_{Y}}$

Here $\mu_{c}$ is the combined mean given by:

$\mu_{c}=\frac{n_{X}\mu_{X}+n_{Y}\mu_{Y}}{n_{X}+n_{Y}}$

It is provided that the sample size is same for both the data sets, i.e. $n_{X} = n_{Y}=n$

Compute the combined mean as follows:

$\mu_{c}=\frac{n_{X}\mu_{X}+n_{Y}\mu_{Y}}{n_{X}+n_{Y}}\\=\frac{(n\times10)+(n\times10)}{n+n}}\\=\frac{20n}{2n}\\ =10$

Compute the combined standard deviation as follows:

$SD_{c} =\sqrt{\frac{n_{X}S^{2}_{X}+n_{2}S^{2}_{Y}+n_{X}(\mu_{X}-\mu_{c})^{2}+n_{Y}(\mu_{Y}-\mu_{c})^{2}}{n_{X}+n_{Y}}}\\=\sqrt{\frac{(n\times1^{2})+(n\times2^{2})+(n(10-10))+(n(10-10))}{n+n}}\\=\sqrt{\frac{n+4n}{2n} } \\=\sqrt{\frac{5n}{2n} } \\=\sqrt{\frac{5}{2}} \\=1.58114$