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3 years ago

X+y=40 0.08x+0.03y=1.7

1 answer:

7 0

SOLVING:

$\begin{cases}&x+y=40\quad\Longrightarrow\ Equation\ 1\\&0.08x+0.03y=1.7\quad\Longrightarrow\ Equation\ 2\end{cases}\\ \\ \\y=40-x\\ \\\textbf{Replace\ "y"\ in\ equation\ 2:}\\ \\0.08x+0.03(40-x)=1.7\\ \\0.08x+1.2-0.03x=1.7\\ \\0.05x+1.2=1.7\\ \\0.05x=1.7-1.2\\ \\0.05x=0.5\\ \\x= \dfrac{0.5}{0.05}\\ \\x=10\quad\to\boxed{\boxed{\spadesuit\ First\ answer.\ \spadesuit}}\\ \\ \\y=40-10\\ \\y=30\quad\to\boxed{\boxed{\spadesuit\ Second\ answer.\ \spadesuit}}$

$\textbf{Testing\ in\ first\ equation:}\\10+30=40\\40=40\ \checkmark\\ \\\textbf{Testing\ in\ second\ equation:}\\0.08(10)+0.03(30)=1.7\\0.8+0.9=1.7\\1.7=1.7\ \checkmark\\ \\\textbf{HOPE\ THIS\ HELPS...!!}$

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What is the inequality of this graph?

Over [174]

Answer: y < 3x -4

Step-by-step explanation:

We know two points of line: (0,-4) and (1,-1)

Therefore, we can find the slope: 3/1.

We know the y intercept.

Dashed line means the inequality is not apart of the region.

y < 3x-4

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3 years ago

A university dean is interested in determining the proportion of students who receive some sort of financial aid. Rather than ex

Olenka [21]

Answer:

The 90% confidence interval that estimate the true proportion of students who receive financial aid is

$0.533 < p < 0.64$

$n = 1789$

Step-by-step explanation:

Considering question a

From the question we are told that

The sample size is n = 200

The number of student that receives financial aid is $k = 118$

Generally the sample proportion is

$\^ p = \frac{k}{n}$

=> $\^ p = \frac{118}{200}$

=> $\^ p = 0.59$

From the question we are told the confidence level is 90% , hence the level of significance is

$\alpha = (100 - 90 ) \%$

=> $\alpha = 0.10$

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.645$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \sqrt{\frac{\^ p (1- \^ p)}{n} }$

=> $E = 1.645 * \sqrt{\frac{0.59 (1- 0.59)}{200} }$

=> $E = 0.057$

Generally 90% confidence interval is mathematically represented as

$\^ p -E < p < \^ p +E$

=> $0.533 < p < 0.64$

Considering question b

From the question we are told that

The margin of error is E = 0.03

From the question we are told the confidence level is 99% , hence the level of significance is

$\alpha = (100 - 99 ) \%$

=> $\alpha = 0.01$

Generally from the normal distribution table the critical value of is

$Z_{\frac{\alpha }{2} } = 2.58$

Generally the sample size is mathematically represented as

$[\frac{Z_{\frac{\alpha }{2} }}{E} ]^2 * \^ p (1 - \^ p )$

=> $n = [\frac{2.58}{0.03} ]^2 * 0.59 (1 - 0.59 )$

=> $n = 1789$

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3 years ago

A baker buys 10 pounds of peaches. She buys a total of 25 peaches. Given that 1 lb = 16 oz, what is the average weight, in ounce

creativ13 [48]

Answer:

the average weight of one peach is 6.4 ounces

Step-by-step explanation:

10lb of peaches

16 ounces in a pound

10 x 16 = 160 ounces

divide 160 (ounces) by 25 (Peaches)

= 6.4 ounces per peach

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3 years ago

The sphere below has a radius of 2.5 inches and an approximate volume of 65.42 cubic inches.

Stells [14]

Part a: The radius of the second sphere is 5 inches.

Part b: The volume of the second sphere is 523.33 in³

Part c; The radius of the third sphere is 1.875 inches.

Part d: The volume of the third sphere is 27.59 in³

Explanation:

Given that the radius of the sphere is 2.5 inches.

Part a: We need to determine the radius of the second sphere.

Given that the second sphere has twice the radius of the given sphere.

Radius of the second sphere = 2 × 2.5 = 5 inches

Thus, the radius of the second sphere is 5 inches.

Part b: we need to determine the volume of the second sphere.

The formula to find the volume of the sphere is given by

$V=\frac{4}{3} \pi r^3$

Substituting $\pi=3.14$ and $r=5$ , we get,

$V=\frac{4}{3} (3.14)(125)$

$V=\frac{1580}{3}$

$V=523.3333 \ in^3$

Rounding off to two decimal places, we have,

$V=523.33 \ in^3$

Thus, the volume of the second sphere is 523.33 in³

Part c: We need to determine the radius of the third sphere

Given that the third sphere has a diameter that is three-fourths of the diameter of the given sphere.

Hence, we have,

Diameter of the third sphere = $\frac{3}{4} (5)=3.75$

Radius of the third sphere = $\frac{3.75}{2} =1.875$

Thus, the radius of the third sphere is 1.875 inches

Part d: We need to determine the volume of the third sphere

The formula to find the volume of the sphere is given by

$V=\frac{4}{3} \pi r^3$

Substituting $\pi=3.14$ and $r=1.875$ , we get,

$V=\frac{4}{3} (3.14)(1.875)^3$

$V=\frac{4}{3} (3.14)(6.59)$

$V=27.5901 \ in^3$

Rounding off to two decimal places, we have,

$V=27.59 \ in^3$

Thus, the volume of the third sphere is 27.59 in³

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3 years ago

Solving Systems of linear equations by graphing. no graphing device.

valentina_108 [34]

Answer: (1,3)

Step-by-step explanation:

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2 years ago