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3 years ago

Help. look at the screenshot below.

Mathematics

1 answer:

enot [183]3 years ago

8 0

Answer:

The third option

Step-by-step explanation:

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emmasim [6.3K]

Answer:

Step-by-step explanation:

Given the dataset 147, 154, 156, 161, 162,

Mean is the sum of the dataset divided by the total number of dataset.

a) Mean = $\sum Xi/N$

$\overline x = \dfrac{147+154+156+161+162}{5}\\ \\\overline x = \dfrac{780}{5}\\ \\\overline x= 156$

b) The formula for calculating the deviation from the mean for each value is expressed as $Xi - \overline X$ where;

Xi is value of each item

xbar is the mean = 156

Mean deviation of 147 = 147-156 = -9

Mean deviation of 154 = 154-156 = -2

Mean deviation of 156 = 156-156 = 0

Mean deviation of 161 = 161-156 = 5

Mean deviation of 162 = 162-156 = 6

c) Sum of the deviations $\sum Xi - \overline X$ = (-9-2+0+5+6)

$\sum Xi - \overline X$ = -11+11

$\sum Xi - \overline X$ = 0

<em>Hence the sum of deviation from the mean is 0</em>

5 0

3 years ago

Does anyone know the surface area for this problem. if so i will mark brainliest.

sveticcg [70]

From the given volume, we know that the ratio is:
$\text {Ratio of yellow eraser to white eraser = } \sqrt[3]{ \dfrac{16}{2} }$

From the given area, we know that the ratio is:
$\text {Ratio of yellow eraser to white eraser = } \sqrt{ \dfrac{52}{x} }$

Equate the two ratio and solve for x:
$\sqrt[3]{ \dfrac{16}{2} } = \sqrt{ \dfrac{52}{x} }$

Cube both sides:
$\dfrac{16}{2} = \bigg(\sqrt{ \dfrac{52}{x} }\bigg)^3$

Square both sides:
$\bigg(\dfrac{16}{2} \bigg)^2 = \bigg( \dfrac{52}{x} }\bigg)^3$

Simplify each term:
$\dfrac{16^2}{2^2} = \dfrac{52^3}{x^3}$

Cross multiply:
$256x^3 = 562432$

Divide both sides by 256:
$x^3 = 2197$

Cube root both sides:
$x = 13$

7 0

4 years ago

kelly made one batchof peanut and pretzal snack mix. how many grams does she need to add to the snack mix to make 2 kilograms

Dominik [7]

2000g
1000g=1kg
Hence 2000g =2kg
Kelly needs to add 2000g to the snack mix to make 2kg.

3 0

3 years ago

Solve the equation on the interval [0,2π). cos^4x=cos^4xcscx

Luba_88 [7]

$\bf cos^4(x)=cos^4(x)csc(x)\\\\
-----------------------------\\\\
cos^4(x)=cos^4(x)\cfrac{1}{sin(x)}\implies cos^4(x)=\cfrac{cos^4(x)}{sin(x)}
\\\\\\
cos^4(x)sin(x)=cos^4(x)\implies cos^4(x)sin(x)-cos^4(x)=0
\\\\\\
cos^4(x)[sin(x)-1]=0\to 
\begin{cases}
cos^4(x)=0\to x=cos^{-1}(0)\\
----------\\
sin(x)-1=0\\
sin(x)=1\to x=sin^{-1}(1)
\end{cases}$

$\bf \measuredangle x = cos^{-1}(0)\implies \measuredangle x = 
\begin{cases}
\frac{\pi }{2}\\\\
\frac{3\pi }{2}
\end{cases}
\\\\\\
\measuredangle x=sin^{-1}(1)\implies \measuredangle x=\frac{\pi }{2}$

8 0

4 years ago

A truck is carrying 102 bushels of sweet potatoes, 101

LUCKY_DIMON [66]

Answer:

306

Step-by-step explanation:

101+102+103=306

7 0

3 years ago