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2 years ago

9

The table shows the time it takes jessica to make rubber band bracelets using a loom. Based on the table how many bracelets can

jessica make in 4.5 hours

2 answers:

denis-greek [22]2 years ago

5 0

Answer:

bro ur in k12 right in in there two arnt u in mrs fleeces class?

Step-by-step explanation:

you have to be cuz i seen ur name... u dont know me but ik you..

,,,,α

]

love history [14]2 years ago

4 0

Answer:

18 rubber band bracelets

Step-by-step explanation:

1 hour = 4 bracelets

4.5 hours = ?

4.5 hours/× 4 bracelets

_______

1 hour/

4.5×4 bracelets = 18 bracelets

In an hour Jessica makes four bracelets and in 4.5hours she makes how many bracelets ,we don't know , obviously in 4.5hours she would make more so if more less will divide.

so 4.5hours over 1hour times 4 bracelets. so hour will cancel hour and now 4.5 times 4 bracelets is equal to 18 bracelets

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AnnZ [28]

Answer:

The average rate of change of $f(x) = x^2 +6x+13$ is 12

The average rate of change of $f(x) =- x^2 +6x+13$ is 10

Step-by-step explanation:

The average rate of change of f(x) over an interval between 2 points (a ,f(a)) and (b ,f(b)) is the slope of the secant line connecting the 2 points.

We can calculate the average rate of change between the 2 points by

$\frac{f(b) - f(a)}{b -a}$ -------------------(1)

(1) The average rate of change of the function $f(x) = x^2 +6x+13$ over the interval 1 ≤ x ≤ 5

f(a) = f(1)

$f(1) = (1)^2 +6(1) + 13$

f(1) =1+6+13

f(a) = 20---------------------(2)

f(b) = f(5)

$f(5) = (5)^2 +6(5)+13$

f(5) = 25 +30 +13

f(5) = 68-----------------------(3)

The average rate of change between (1 ,20) and (5 ,68 ) is

Substituting eq(2) and(3) in (1)

= $\frac{f(5) - f(1)}{5-1}$

= $\frac{68 -20}{5-1}$

= $\frac{48}{4}$

=12

This means that the average of all the slopes of lines tangent to the graph of f(x) between (1 ,20) and (5 ,68 ) is 12

(2) The average rate of change of the function $f(x) = -x^2 +6x+13$ over the interval -1 ≤ x ≤ 5

f(a) = f(-1)

$f(1) = (-1)^2 +6(-1) + 13$

f(1) =1-6+13

f(1) = 8---------------------(4)

f(b) = f(5)

$f(5) = (5)^2 +6(5)+13$

f(5) = 25 +30 +13

f(5) = 68-----------------------(5)

The average rate of change between (-1 ,8) and (5 ,68 ) is

Equation (1) becomes

$\frac{f(5) - f(-1)}{5-(-1)}$

On substituting the values

= $\frac{68 - 8}{5-(-1)}$

= $\frac{60}{5+1}$

= $\frac{60}{6}$

= 10

This means that the average of all the slopes of lines tangent to the graph of f(x) between (-1 ,8) and (5 ,68 ) is 10

4 0

3 years ago

Solid fats are more likely to raise blood cholesterol levels than liquid fats. Suppose a nutritionist analyzed the percentage of

Andrews [41]

Answer:

$t = 31.29$

Step-by-step explanation:

Given

$\begin{array}{ccccccc}{Stick} & {25.8} & {26.9} & {26.2} & {25.3} & {26.7}& {26.1} \ \\ {Liquid} & {16.9} & {17.4} & {16.8} & {16.2} & {17.3}& {16.8} \ \end{array}$

Required

Determine the test statistic

Let the dataset of stick be A and Liquid be B.

We start by calculating the mean of each dataset;

$\bar x =\frac{\sum x}{n}$

n, in both datasets in 6

For A

$\bar x_A =\frac{25.8+26.9+26.2+25.3+26.7+26.1}{6}$

$\bar x_A =\frac{157}{6}$

$\bar x_A =26.17$

For B

$\bar x_B =\frac{16.9+17.4+16.8+16.2+17.3+16.8}{6}$

$\bar x_B =\frac{101.4}{6}$

$\bar x_B =16.9$

Next, calculate the sample standard deviation

This is calculated using:

$s = \sqrt{\frac{\sum(x - \bar x)^2}{n-1}}$

For A

$s_A = \sqrt{\frac{\sum(x - \bar x_A)^2}{n-1}}$

$s_A = \sqrt{\frac{(25.8-26.17)^2+(26.9-26.17)^2+(26.2-26.17)^2+(25.3-26.17)^2+(26.7-26.17)^2+(26.1-26.17)^2}{6-1}}$

$s_A = \sqrt{\frac{1.7134}{5}}$

$s_A = \sqrt{0.34268}$

$s_A = 0.5854$

For B

$s_B = \sqrt{\frac{\sum(x - \bar x_B)^2}{n-1}}$

$s_B = \sqrt{\frac{(16.9 - 16.9)^2+(17.4- 16.9)^2+(16.8- 16.9)^2+(16.2- 16.9)^2+(17.3- 16.9)^2+(16.8- 16.9)^2}{6-1}}$

$s_B = \sqrt{\frac{0.92}{5}}$

$s_B = \sqrt{0.184}$

$s_B = 0.4290$

Calculate the pooled variance

$S_p^2 = \frac{(n_A - 1)*s_A^2 + (n_B - 1)*s_B^2}{(n_A+n_B-2)}$

$S_p^2 = \frac{(6 - 1)*0.5854^2 + (6 - 1)*0.4290^2}{(6+6-2)}$

$S_p^2 = \frac{2.6336708}{10}$

$S_p^2 = 0.2634$

Lastly, calculate the test statistic using:

$t = \frac{(\bar x_A - \bar x_B) - (\mu_A - \mu_B)}{\sqrt{S_p^2/n_A +S_p^2/n_B}}$

We set

$\mu_A = \mu_B$

So, we have:

$t = \frac{(\bar x_A - \bar x_B) - (\mu_A - \mu_A)}{\sqrt{S_p^2/n_A +S_p^2/n_B}}$

$t = \frac{(\bar x_A - \bar x_B) }{\sqrt{S_p^2/n_A +S_p^2/n_B}}$

The equation becomes

$t = \frac{(26.17 - 16.9) }{\sqrt{0.2634/6 +0.2634/6}}$

$t = \frac{9.27}{\sqrt{0.0878}}$

$t = \frac{9.27}{0.2963}$

$t = 31.29$

<em>The test statistic is 31.29</em>

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2 years ago

Find an equation of the line satisfying the given conditions. horizontal; through (-3, -8)

Irina18 [472]

Y = -8
y is equal to -8

7 0

3 years ago

Write add, divide, multiply, and subtract in the correct order to complete the following sentence.

Bad White [126]

Answer:

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5 0

2 years ago

When using "mean" does it mean average of something if not please explain, I'm a bit confused.

lesya [120]

Mean the the total of all values, divided by the number of values

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2 years ago

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