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Cerrena [4.2K]
3 years ago
5

Can anyone help me with these?

Mathematics
1 answer:
katrin2010 [14]3 years ago
6 0

Answer:

Step-by-step explanation:

Distributive property: a*(b + c)  = ab + ac

16)5x²(7 - x²) = 5x²*7 - 5x²*x²

                   =35x² - 5x⁴

17)

10x(4 - 3x) = 10x*4 - 10x*3x

                  = 40x - 30x²

18) 9( y +4) = 9*y + 9*4

                 = 9y + 36

19) 3(4 - x) - 3(2 - x) = 3*4 - 3*x -3*2 - x*(-3)

                                = 12 - 3x - 6  + 3x  {add/subtract like terms}

                                = 12 - 6 - 3x + 3x

                                = 6

20) 6y + 6(y - 4) = 6y + 6*y - 6*4

                          = 6y + 6y - 24

                          = 12y - 24

                         

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The following table shows scores obtained in an examination by B.Ed JHS Specialism students. Use the information to answer the q
Makovka662 [10]

Answer:

(a) The cumulative frequency curve for the data is attached below.

(b) (i) The inter-quartile range is 10.08.

(b) (ii) The 70th percentile class scores is 0.

(b) (iii) the probability that a student scored at most 50 on the examination is 0.89.

Step-by-step explanation:

(a)

To make a cumulative frequency curve for the data first convert the class interval into continuous.

The cumulative frequencies are computed by summing the previous frequencies.

The cumulative frequency curve for the data is attached below.

(b)

(i)

The inter-quartile range is the difference between the third and the first quartile.

Compute the values of Q₁ and Q₃ as follows:

Q₁ is at the position:

\frac{\sum f}{4}=\frac{100}{4}=25

The class interval is: 34.5 - 39.5.

The formula of first quartile is:

Q_{1}=l+[\frac{(\sum f/4)-(CF)_{p}}{f}]\times h

Here,

l = lower limit of the class consisting value 25 = 34.5

(CF)_{p} = cumulative frequency of the previous class = 24

f = frequency of the class interval = 20

h = width = 39.5 - 34.5 = 5

Then the value of first quartile is:

Q_{1}=l+[\frac{(\sum f/4)-(CF)_{p}}{f}]\times h

     =34.5+[\frac{25-24}{20}]\times5\\\\=34.5+0.25\\=34.75

The value of first quartile is 34.75.

Q₃ is at the position:

\frac{3\sum f}{4}=\frac{3\times100}{4}=75

The class interval is: 44.5 - 49.5.

The formula of third quartile is:

Q_{3}=l+[\frac{(3\sum f/4)-(CF)_{p}}{f}]\times h

Here,

l = lower limit of the class consisting value 75 = 44.5

(CF)_{p} = cumulative frequency of the previous class = 74

f = frequency of the class interval = 15

h = width = 49.5 - 44.5 = 5

Then the value of third quartile is:

Q_{3}=l+[\frac{(3\sum f/4)-(CF)_{p}}{f}]\times h

     =44.5+[\frac{75-74}{15}]\times5\\\\=44.5+0.33\\=44.83

The value of third quartile is 44.83.

Then the inter-quartile range is:

IQR = Q_{3}-Q_{1}

        =44.83-34.75\\=10.08

Thus, the inter-quartile range is 10.08.

(ii)

The maximum upper limit of the class intervals is 69.5.

That is the maximum percentile class score is 69.5th percentile.

So, the 70th percentile class scores is 0.

(iii)

Compute the probability that a student scored at most 50 on the examination as follows:

P(\text{Score At most 50})=\frac{\text{Favorable number of cases}}{\text{Total number of cases}}

                                 =\frac{10+4+10+20+30+15}{100}\\\\=\frac{89}{100}\\\\=0.89

Thus, the probability that a student scored at most 50 on the examination is 0.89.

5 0
3 years ago
Which of the pairs of events below is mutually exclusive?
Nady [450]

Answer:

Drawing an ace of spades and then drawing another ace of spades without replacement from a standard deck of cards.

Step-by-step explanation:

Given four pair of events:

1. Drawing an ace of spades and then drawing another ace of spades without replacement from a standard deck of cards

2. Drawing a 2 and drawing a 4 with replacement from a standard deck of cards

3. Drawing a heart and then drawing a spade without replacement from a standard deck of cards

4. Drawing a jack and then drawing a 7 without replacement from a standard deck of cards

To find:

Which of the pairs is mutually exclusive?

Solution:

First of all, let us have a look at the definition of mutually exclusive events.

Mutually exclusive events do not have any case in common.

In other words, two events are known as mutually exclusive when both the events can not occur at the same time.

Now, let us have a look at the given options one by one:

1. Drawing an ace of spades and then drawing another ace of spades without replacement from a standard deck of cards.

We know that there is only one ace of spades in a standard deck.

Therefore, the two events can not occur one after the other, so these are mutually exclusive events.

2. Drawing a 2 and drawing a 4 with replacement from a standard deck of cards.

The two events can occur one after the other, so these are not mutually exclusive.

3. Drawing a heart and then drawing a spade without replacement from a standard deck of cards.

Separate cards are drawn one after the other, so not mutually exclusive.

4. Drawing a jack and then drawing a 7 without replacement from a standard deck of cards.

Separate cards are drawn one after the other, so not mutually exclusive.

3 0
3 years ago
$900,000 is what percent of $900,000?
USPshnik [31]
What type of question is this
8 0
3 years ago
Read 2 more answers
What is the radius of a circle that has a diameter of 199
wolverine [178]

Answer:

99.5

Step-by-step explanation:

Radius = Diameter/2

199/9

=99.5metres

6 0
2 years ago
I need help math please
iVinArrow [24]
Positive linear association !
6 0
3 years ago
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