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1 year ago

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Maya's chocolate bar is 29% cocoa. If the weight of the chocolate bar is 67 grams, how many grams of cocoa does it contain? Roun

d your answer to the nearest tenth.

1 answer:

nika2105 [10]1 year ago

8 0

Answer:

20

Step-by-step explanation:

x/67=29/100

29x67=1943

1943/100=

19.43

Round:

20

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Complete the equation of the line through (-9,-9) and (-6,0)

Rudik [331]

For this case we have that by definition, the equation of the line of the slope-intersection form is given by:

$y = mx + b$

Where:

m: It is the slope of the line

b: It is the cut-off point with the y axis.

According to the data of the statement we have the following points:

$(x_ {1}, y_ {1}): (- 9, -9)\\(x_ {2}, y_ {2}): (- 6,0)$

We found the slope:

$m = \frac {y_ {2} -y_ {1}} {x_ {2} -x_ {1}} = \frac {0 - (- 9)} {- 6 - (- 9)} = \frac { 9} {- 6 + 9} = \frac {9} {3} = 3$

Thus, the equation is of the form:

$y = 3x + b$

We substitute one of the points and find b:

$0 = 3 (-6) + b\\0 = -18 + b\\b = 18$

Finally, the equation is:

$y = 3x + 18$

Answer:

$y = 3x + 18$

7 0

3 years ago

Which expression is equivalent to 5!<br> 5^4*3^2<br> 5-4-3-2-1<br> 5.4.3-2.1<br> 5+4+3+2+1<br> 5.1

baherus [9]

The first one is like 5x5x5x5 which is 625 multiplied by 3x3 which is 9 and together the product will be 5,625.

The second one is 5-4 which is 1-3 which is -2 - 2 = -4 - 1 which is -5 so it’s not it because it’s the opposite of the number we need to find.

The third one is kind of messed up because there’s no such thing as 5.4.3 so sorry I can’t do that one.

The fourth one is 15 because you basically add the numbers up.

And the fifth is not 5. It’s 5.1 so that’s not it either.

So I’m sorry I can’t find the answer but at least I have you some background :/

6 0

3 years ago

What can you predict regarding the probability of the

natka813 [3]

Answer:

The frequency does not change with more trials

Step-by-step explanation:

To predict: the probability of the coin landing heads up

Solution:

Probability refers to the chances that an event will occur in an experiment. The value of probability lies between 0 and 1. 0 indicates impossible event and 1 indicates a sure event. The probability of an event can not be greater than 1.

When a coin is tossed, there are two possible outcomes: heads (H), tails (T).

In case of the probability of the coin landing heads up, the frequency does not change with more trials.

6 0

3 years ago

Read 2 more answers

Help I really don’t understand this

MAXImum [283]

Let x=3.77777777...
10x=37.777777777...
9x=10x-x=34
9x=34 so x=34/9=3 7/9

3 0

2 years ago

Activity 4: Performance Task

Nookie1986 [14]

An arithmetic progression is simply a progression with a common difference among consecutive terms.

<em>The sum of multiplies of 6 between 8 and 70 is 390</em>
<em>The sum of multiplies of 5 between 12 and 92 is 840</em>
<em>The sum of multiplies of 3 between 1 and 50 is 408</em>
<em>The sum of multiplies of 11 between 10 and 122 is 726</em>
<em>The sum of multiplies of 9 between 25 and 100 is 567</em>
<em>The sum of the first 20 terms is 630</em>
<em>The sum of the first 15 terms is 480</em>
<em>The sum of the first 32 terms is 3136</em>
<em>The sum of the first 27 terms is -486</em>
<em>The sum of the first 51 terms is 2193</em>

<em />

<u>(a) Sum of multiples of 6, between 8 and 70</u>

There are 10 multiples of 6 between 8 and 70, and the first of them is 12.

This means that:

$\mathbf{a = 12}$

$\mathbf{n = 10}$

$\mathbf{d = 6}$

The sum of n terms of an AP is:

$\mathbf{S_n = \frac n2(2a + (n - 1)d)}$

Substitute known values

$\mathbf{S_{10} = \frac{10}2(2*12 + (10 - 1)6)}$

$\mathbf{S_{10} = 390}$

<u>(b) Multiples of 5 between 12 and 92</u>

There are 16 multiples of 5 between 12 and 92, and the first of them is 15.

This means that:

$\mathbf{a = 15}$

$\mathbf{n = 16}$

$\mathbf{d = 5}$

The sum of n terms of an AP is:

$\mathbf{S_n = \frac n2(2a + (n - 1)d)}$

Substitute known values

$\mathbf{S_{16} = \frac{16}2(2*15 + (16 - 1)5)}$

$\mathbf{S_{16} = 840}$

<u>(c) Multiples of 3 between 1 and 50</u>

There are 16 multiples of 3 between 1 and 50, and the first of them is 3.

This means that:

$\mathbf{a = 3}$

$\mathbf{n = 16}$

$\mathbf{d = 3}$

The sum of n terms of an AP is:

$\mathbf{S_n = \frac n2(2a + (n - 1)d)}$

Substitute known values

$\mathbf{S_{16} = \frac{16}2(2*3 + (16 - 1)3)}$

$\mathbf{S_{16} = 408}$

<u>(d) Multiples of 11 between 10 and 122</u>

There are 11 multiples of 11 between 10 and 122, and the first of them is 11.

This means that:

$\mathbf{a = 11}$

$\mathbf{n = 11}$

$\mathbf{d = 11}$

The sum of n terms of an AP is:

$\mathbf{S_n = \frac n2(2a + (n - 1)d)}$

Substitute known values

$\mathbf{S_{16} = \frac{11}2(2*11 + (11 - 1)11)}$

$\mathbf{S_{11} = 726}$

<u />

<u>(e) Multiples of 9 between 25 and 100</u>

There are 9 multiples of 9 between 25 and 100, and the first of them is 27.

This means that:

$\mathbf{a = 27}$

$\mathbf{n = 9}$

$\mathbf{d = 9}$

The sum of n terms of an AP is:

$\mathbf{S_n = \frac n2(2a + (n - 1)d)}$

Substitute known values

$\mathbf{S_{9} = \frac{9}2(2*27 + (9 - 1)9)}$

$\mathbf{S_{9} = 567}$

<u>(f) Sum of first 20 terms</u>

The given parameters are:

$\mathbf{a = 3}$

$\mathbf{d = 3}$

$\mathbf{n = 20}$

The sum of n terms of an AP is:

$\mathbf{S_n = \frac n2(2a + (n - 1)d)}$

Substitute known values

$\mathbf{S_{20} = \frac{20}2(2*3 + (20 - 1)3)}$

$\mathbf{S_{20} = 630}$

<u>(f) Sum of first 15 terms</u>

The given parameters are:

$\mathbf{a = 4}$

$\mathbf{d = 4}$

$\mathbf{n = 15}$

The sum of n terms of an AP is:

$\mathbf{S_n = \frac n2(2a + (n - 1)d)}$

Substitute known values

$\mathbf{S_{15} = \frac{15}2(2*4 + (15 - 1)4)}$

$\mathbf{S_{15} = 480}$

<u>(g) Sum of first 32 terms</u>

The given parameters are:

$\mathbf{a = 5}$

$\mathbf{d = 6}$

$\mathbf{n = 32}$

The sum of n terms of an AP is:

$\mathbf{S_n = \frac n2(2a + (n - 1)d)}$

Substitute known values

$\mathbf{S_{32} = \frac{32}2(2*5 + (32 - 1)6)}$

$\mathbf{S_{32} = 3136}$

<u>(g) Sum of first 27 terms</u>

The given parameters are:

$\mathbf{a = 8}$

$\mathbf{d = -2}$

$\mathbf{n = 27}$

The sum of n terms of an AP is:

$\mathbf{S_n = \frac n2(2a + (n - 1)d)}$

Substitute known values

$\mathbf{S_{27} = \frac{27}2(2*8 + (27 - 1)*-2)}$

$\mathbf{S_{27} = -486}$

<u>(h) Sum of first 51 terms</u>

The given parameters are:

$\mathbf{a = -7}$

$\mathbf{d = 2}$

$\mathbf{n = 51}$

The sum of n terms of an AP is:

$\mathbf{S_n = \frac n2(2a + (n - 1)d)}$

Substitute known values

$\mathbf{S_{51} = \frac{51}2(2*-7 + (51 - 1)*2)}$

$\mathbf{S_{51} = 2193}$

Read more about arithmetic progressions at:

brainly.com/question/13989292

4 0

1 year ago

Read 2 more answers