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3 years ago

6

The Bengals football team scored 2 touchdowns for 6 points each, one extra point, and 3 field goals for 3 points each. The Raven

s scored an equal number of points, except for an additional field goal. What was the final score for the game?

1 answer:

Yanka [14]3 years ago

4 0

bangles scored 22 pts but don't know how much the ravens scored

You might be interested in

a) How many three-digit numbers can be formed from the digits 0, 1, 2, 3, 4, 5, and 6?6x7x7=294 b) How many three-digit numbers

love history [14]

Answer:

a) 294

b) 180

c) 75

d) 168

e) 105

Step-by-step explanation:

Given the numbers 0, 1, 2, 3, 4, 5 and 6.

Part A)

How many 3 digit numbers can be formed ?

Solution:

Here we have 3 spaces for the digits.

Unit's place, ten's place and hundred's place.

For unit's place, any of the numbers can be used i.e. 7 options.

For ten's place, any of the numbers can be used i.e. 7 options.

For hundred's place, 0 can not be used (because if 0 is used here, the number will become 2 digit) i.e. 6 options.

Total number of ways = $7 \times 7 \times 6$ = <em>294 </em>

<em></em>

<em>Part B:</em>

How many 3 digit numbers can be formed if repetition not allowed?

Solution:

Here we have 3 spaces for the digits.

Unit's place, ten's place and hundred's place.

For hundred's place, 0 can not be used (because if 0 is used here, the number will become 2 digit) i.e. 6 options.

Now, one digit used, So For unit's place, any of the numbers can be used i.e. 6 options.

Now, 2 digits used, so For ten's place, any of the numbers can be used i.e. 5 options.

Total number of ways = $6 \times 6 \times 5$ = <em>180</em>

<em></em>

<em>Part C)</em>

How many odd numbers if each digit used only once ?

Solution:

For a number to be odd, the last digit must be odd i.e. unit's place can have only one of the digits from 1, 3 and 5.

Number of options for unit's place = 3

Now, one digit used and 0 can not be at hundred's place So For hundred's place, any of the numbers can be used i.e. 5 options.

Now, 2 digits used, so For ten's place, any of the numbers can be used i.e. 5 options.

Total number of ways = $3 \times 5 \times 5$ = <em>75</em>

<em></em>

<em>Part d)</em>

How many numbers greater than 330 ?

Case 1: 4, 5 or 6 at hundred's place

Number of options for hundred's place = 3

Number of options for ten's place = 7

Number of options for unit's place = 7

Total number of ways = $3 \times 7 \times 7$ = 147

Case 2: 3 at hundred's place

Number of options for hundred's place = 1

Number of options for ten's place = 3 (4, 5, 6)

Number of options for unit's place = 7

Total number of ways = $1 \times 3 \times 7$ = 21

Total number of required ways = 147 + 21 = <em>168</em>

<em></em>

<em>Part e)</em>

Case 1: 4, 5 or 6 at hundred's place

Number of options for hundred's place = 3

Number of options for ten's place = 6

Number of options for unit's place = 5

Total number of ways = $3 \times 6 \times 5$ = 90

Case 2: 3 at hundred's place

Number of options for hundred's place = 1

Number of options for ten's place = 3 (4, 5, 6)

Number of options for unit's place = 5

Total number of ways = $1 \times 3 \times 5$ = 15

Total number of required ways = 90 + 15 = <em>105</em>

7 0

3 years ago

1011+111 in binary form

Ann [662]

Answer:

10010

Step-by-step explanation:

$1011=1(2)^3+0(2)^2+1(2)^1+1(2)^0$

$111=1(2)^2+1(2)^1+1(2)^0$

So $1011+111$ gives us:

$1(2)^3+0(2)^2+1(2)^1+1(2)^0$

$+$

$1(2)^2+1(2)^1+1(2)^0$

-----------------------------------------------------

Combine like terms:

$1(2)^3+(0+1)(2)^2+(1+1)(2)^1+(1+1)(2)^0$

$1(2)^3+1(2)^2+(2)(2)^1+(2)(2)^0$

We aren't allowed to have a coefficient bigger than 1.

I'm going to replace $2^0$ with 1 and $2$ with $(2)^1$ :

$1(2)^3+1(2)^2+(2)^2+(2)^1(1)$

I want a $2^0$ number:

$1(2)^3+1(2)^2+1(2)^2+1(2)^1+0(2)^0$

Combine like terms:

$1(2)^3+2(2)^2+1(2)^1+0(2)^0$

$2(2)^2=2^3$ :

$1(2)^3+2^3+1(2)^1+0(2)^0$

Combine like terms:

$2(2)^3+1(2)^1+0(2)^0$

We can rewrite the first term by law of exponents:

$2^4+1(2)^1+0(2)^0$

$1(2)^4+1(2)^1+0(2)^0$

So the binary form is:

$10010$

Maybe you like this way more:

Keep in mind 1+1=10 and that 1+1+1=11:

Setup:

1 0 1 1

+ 1 1 1

------------------------------

(1) (1) (1)

1 0 1 1

+ 1 1 1

------------------------------

1 0 0 1 0

I had to do some carry over with my 1+1=10 and 1+1+1=11.

8 0

3 years ago

Eric bought a canoe for $333.75. The sales tax was 5.25%. What was his total cost?

yan [13]

Answer:

333.80

Step-by-step explanation:

3 0

3 years ago

Match each pair of monomials with their greatest common factor. Tiles 4p4q5 and 8p3q2 24pq6 and 16p3q5 8p3q6 and 4p2q5 16p3q2 an

pashok25 [27]

The GCF of the first two is 4p³q².The GCF of the second two is 8pq⁵.The GCF of the third two is 4p²q⁵.The GCF of the fourth two is 8p²q.The GCF of the fifth two is 4p²q.
To find the GCF of each pair, find the greatest number that will divide into each coefficient. As for the variable portions, choose the variable that has the smallest exponent from each pair.

8 0

3 years ago

What fraction of a circle measures 20 deegres

Temka [501]

A circle is 360 degrees so 360 divided by 20=18 so 20 degrees is one 18th of a circle

5 0

3 years ago

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