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3 years ago

Type the correct answer in the box. Use numerals instead of words.

Mathematics

1 answer:

KIM [24]3 years ago

3 0

Answer:

y = -.05x + 6

Step-by-step explanation:

Plug in your x and y coordinates

1 = -.05(10) + b

1 = -5 + b

6 = b

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This we havent learned but teacher gave us for end of year est plz help

chubhunter [2.5K]

Answer:

Step-by-step explanation:

Finding total number of pupils(y)

3/y x 360°=27°

3x360°/y=27°

1080°/y=27°

1080°=27°y

1080°/27°=27°y/27°

40=y

Number of pupils who said that Maths was their favorite (z)

z/40x360°=135°

360°z/40=135°

360°z=135°x40

360°z=5400°

360°z/360°=5400°/360°

z=15

3 0

2 years ago

J is building a treehouse. He wants the treehouse to be 3 meters long. If j builds 50 centimeters. Each dayhow many days will it

Allushta [10]

50 cm = 1/2 m
3 m ÷ 1/2 m = 6

It will take him six days.

3 0

3 years ago

Five students visiting the student health center for a free dental examination during National Dental Hygiene Month were asked h

kirza4 [7]

Answer:

95% confidence interval for the mean number of months is between a lower limit of 6.67 months and an upper limit of 25.73 months.

Step-by-step explanation:

Confidence interval is given as mean +/- margin of error (E)

Data: 5, 15, 12, 22, 27

mean = (5+15+12+22+27)/5 = 81/5 = 16.2 months

sd = sqrt[((5-16.2)^2 + (15-16.2)^2 + (12-16.2)^2 + (22-16.2)^2 + (27-16.2)^2) ÷ 5] = sqrt(58.96) = 7.68 months

n = 5

degree of freedom = n-1 = 5-1 = 4

confidence level (C) = 95% = 0.95

significance level = 1 - C = 1 - 0.95 = 0.05 = 5%

critical value (t) corresponding to 4 degrees of freedom and 5% significance level is 2.776

E = t×sd/√n = 2.776×7.68/√5 = 9.53 months

Lower limit of mean = mean - E = 16.2 - 9.53 = 6.67 months

Upper limit of mean = mean + E = 16.2 + 9.53 = 25.73 months

95% confidence interval is (6.67, 25.73)

3 0

2 years ago

I need help with #11

Troyanec [42]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\

\begin{array}{rllll} 
% left side templates
f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
y=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}}
\end{array}$

$\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}
\\\\
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\
\end{array}$

$\bf \begin{array}{llll}


\bullet \textit{ vertical shift by }{{ D}}\\
\qquad if\ {{ D}}\textit{ is negative, downwards}\\\\
\qquad if\ {{ D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{ B}}}
\end{array}$

now, with that template in mind, let's take a peek at this function

$\bf \begin{array}{lllcclll}
y=&2(&1x&-2)^2&-4\\
&\uparrow &\uparrow &\uparrow &\uparrow \\
&A&B&C&D
\end{array}\\\\
-----------------------------\\\\
A\cdot B=2\impliedby \textit{shrunk by a factor of 2, of half-size}\\\\
\cfrac{C}{B}= \cfrac{-2}{1}\implies -2\impliedby \textit{horizontal right shift of 2 units}\\\\
D=-4\impliedby \textit{vertical down shift of 4 units}$

so, the graph of y=2(x-2)²-4, is really the same graph of y=x², BUT, narrower, and moved about horizontally and vertically

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This line graph shows the temperature in London over a period of time.

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The temperature in London at 2pm is 14°C

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