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2 years ago

A spider walks on the outside of a box from point A to B to C to D

Mathematics

1 answer:

Nikitich [7]2 years ago

6 0

The path consists of 4 line segments. Each line segment is the hypotenuse of a different right triangle.

length of line segment AB = $\sqrt{2^{2}+5^{2} } =\sqrt{29} =5.39cm$

length of line segment BC = $\sqrt{3^{2}+4^{2} } =\sqrt{25} =5cm$

length of line segment CD = $\sqrt{3^{2}+5^{2} } =\sqrt{34} =5.83cm$

length of line segment DE $=\sqrt{2.5^{2}+3^{2} } } =\sqrt{15.25} =3.91cm$

The total length of the spider's path is about 20.13 cm.

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Answer:

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Step-by-step explanation:

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2 years ago

Calculate the sample standard deviation and sample variance for the following frequency distribution of hourly wages for a sampl

lidiya [134]

Answer:

(a) The sample variance is 16.51

(a) The sample standard deviation is 4.06

Step-by-step explanation:

Given

$\begin{array}{cc}{Class} & {Frequency} & 8.26 - 10.00 & 20 &10.01-11.75 & 38 &11.76 - 13.50& 36 & 13.51-15.25 &25&15.26-17.00 &27 &\ \end{array}$

Solving (a); The sample variance.

First, calculate the class midpoints.

This is the mean of the intervals.

i.e.

$x_1 = \frac{8.26+10.00}{2} = \frac{18.26}{2} = 9.13$

$x_2 = \frac{10.01+11.75}{2} = \frac{21.76}{2} = 10.88$

$x_3 = \frac{11.76+13.50}{2} = \frac{25.26}{2} = 12.63$

$x_4 = \frac{13.51+15.25}{2} = \frac{28.76}{2} = 14.38$

$x_5 = \frac{15.26+17.00}{2} = \frac{32.26}{2} = 16.13$

So, the table becomes:

$\begin{array}{ccc}{Class} & {Frequency} & {x} & 8.26 - 10.00 & 20&9.13 &10.01-11.75 & 38 &10.88&11.76 - 13.50& 36 &12.63& 13.51-15.25 &25&14.38&15.26-17.00 &27 &16.13\ \end{array}$

Next, calculate the mean

$\bar x = \frac{\sum fx}{\sum f}$

$\bar x = \frac{20*9.13 + 38 * 10.88+36*12.63+25*14.38+27*16.13}{20+38+36+25+27}$

$\bar x = \frac{1845.73}{146}$

$\bar x = 12.64$

Next, the sample variance is:

$\sigma^2 = \frac{\sum f(x - \bar x)^2}{\sum f - 1}$

So, we have:

$\sigma^2 = \frac{20*(9.13-12.63)^2 + 38 * (10.88-12.63)^2 +...........+27 * (16.13 -12.63)^2}{20+38+36+25+27-1}$

$\sigma^2 = \frac{2393.6875}{145}$

$\sigma^2 = 16.51$

The sample standard deviation is:

$\sigma = \sqrt{\sigma^2}$

$\sigma = \sqrt{16.51}$

$\sigma = 4.06$

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2 years ago

The table shows the relationship between the number of almonds and the number of cashews in various measures of chocolate fudge.

katovenus [111]

Answer:

A = 9

Step-by-step explanation:

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2 years ago

The table below represents a linear function f(x) and the equation represents a function g(x):

Aneli [31]

Slope of f(x)

Slope of g(x)

PART A)
The slopes of f(x) and g(x) are the same.

PART B)
The equation for line f(x)
y = 4x + b
-1 = 4(0) + b
-1 = b

y = 4x - 1

The function G(x) has a bigger y-intercept. The y-intercept of g(x) is 3, while the y-intercept of f(x) is -1.

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3 years ago

Bryce tried to solve an equation step by step. \qquad\begin{aligned} \dfrac83&=3\left(c+\dfrac53\right)\\\\ \\ \dfrac83&

sesenic [268]

Answer:

Bryce is wrong in step 1 because he did not distribute 3 over 5/3

Explanation

Given the steps taken by bryce as shown, we are to find where he made an error

$\qquad\begin{aligned} \dfrac83&=3\left(c+\dfrac53\right)\\\\ \\ \dfrac83&=3c+\dfrac53&\green{\text{Step } 1}\\\\ \\ 1&=3c&\blue{\text{Step } 2}\\\\ \\ \dfrac13&=c&\purple{\text{Step } 3}\\\\ \end{aligned}$

Given the expression;

$\dfrac83&=3\left(c+\dfrac53\right)\\\\ \\$

Step 1:Expand the bracket using the distributive law;

8/3 = 3c + 3(5/3)

<em>Simplify</em>

8/3 = 3c + 15/3

Step 2: Subtract 15/3 from both sides

8/3 - 15/3 = 3c+15/3-15/3

(8-15)/3 = 3c

-7/3 = 3c

Step 3: Multiply both sides by 1/3

-7/3 * 1/3 = 3c * 1/3

-7/9 = c

Swap

c = -7/9

From the calculation, we can see that Bryce is wrong in step 1 because he did not distribute 3 over 5/3 thereby making his solution incorrect

3 0

2 years ago