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Gre4nikov [31]
3 years ago
6

The actual length of a soccer field is 300 feet. If the width of the field is 180 feet, then what should the width of the scale

drawing be in inches?
Mathematics
1 answer:
topjm [15]3 years ago
6 0

Answer:

2160 inches

Step-by-step explanation:

The actual length of a soccer field is 300 feet. If the width of the field is 180 feet, then what should the width of the scale drawing be in inches?

1 foot = 12 inches

Width of the field = 180 feet

Hence:

1 foot = 12 inches

180 feet = x

x = 180 × 12 inches

x = 2160 inches

Width of the scale drawing in inches = 2160 inches

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Please help me with this problem thank you
horsena [70]

Answer:

take a picture of it and then put it up

Step-by-step explanation:

and they should give you the answer

5 0
3 years ago
Read 2 more answers
If you roll two fair dice repeatedly, what is the probability that you will get a sum of 4 before you get a sum of 5 ? (a) (b) (
AlexFokin [52]

Answer:

The probability that you will get a sum of 4 before you get a sum of 5 is \frac{1}{12}

Step-by-step explanation:

Given : If you roll two fair dice repeatedly.

To find : What is the probability that you will get a sum of 4 before you get a sum of 5 ?

Solution :

When two dice are rolled the outcomes are

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

Total number of outcomes = 36

Favorable outcome get a sum of 4 before you get a sum of 5 is (1,3) ,(2,2) and (3,1) = 3

The probability that you will get a sum of 4 before you get a sum of 5 is

P=\frac{3}{36}

P=\frac{1}{12}

Therefore, The probability that you will get a sum of 4 before you get a sum of 5 is \frac{1}{12}

5 0
3 years ago
The data list shows the scores of ten students in Mr. Smith's math class. 61, 67, 81, 83, 87, 88, 89, 90, 98, 100 What is the st
cluponka [151]

Answer:

12.3

Step-by-step explanation:

Step 1

We find the mean

The data list shows the scores of ten students in Mr. Smith's math class. 61, 67, 81, 83, 87, 88, 89, 90, 98, 100

Mean = Sum of terms/Number of terms

Number of terms = 10

Mean = 61 + 67 + 81 + 83 + 87 + 88 + 89 + 90 + 98 + 100/10

Mean = 844/10

Mean = 84.4

Step 2

Standard deviation

The formula for sample standard deviation =

√(x - Mean)²/n - 1

= √[(61 - 84.4)² + (67 - 84.4)² + (81 - 84.4)² + (83 - 84.4)² + (87 - 84.4)² + (88 - 84.4)² + (89 - 84.4)² + (90 - 84.4)² + (98 - 84.4)² + (100 - 84.4)²]/10 - 1

=√ 547.56 + 302.76 + 11.56 + 1.96 + 6.76 + 12.96 + 21.16 + 31.36 + 184.96 + 243.36/10 - 1

= √1364.4/9

= √151.6

= 12.31259518

Approximately to the nearest tenth = 12.3

The standard deviation = 12.3

6 0
2 years ago
i have been trying to solve this compound and double angle question please help me find the answer to these question guys​
natali 33 [55]

Answer:

These type of questions are super tricky b/c you have to remember all the different versions of the identities, and then they put the question in some odd form,  I feel like this should land math professors in jail , for dishonesty , b/c it's really a form of "how tricky can I make a question and still have a way to solve it"   anyway,

Step-by-step explanation:

a)

next the question asks   1-cos 2A   and this is total abuse of notation.   the way this should be written is   1- cos( 2A)  so we know that the A is part of the cosine functions input... btw.. in any computer program,  it would never ever let you get away with that top form of the expression.  :/   anyway... I keep ranting.. huh... sorry  :P

1-cos(2A) is an odd form of the identity  1/2(1-cos(2A) = sin^{2}(A)  the 1/2 is missing but we can add that pretty easy, we just have to remember to take it out too. I usually forget to do that. and my professor marks me off completely,  totally wrong, but I just miss one small thing  :/  anyway....

our 1-cos(2A) needs the 1/2 added to it.  or if we move that 1/2 to the other side it looks like  2*sin^{2}(A)  = 1-cos(2A)  and this is that "odd" from of the identity that I was talking about.  

next let's deal with sin(2A)  it has an identity of  2 sin(A)cos(A) which is really nice for us b/c it will cancel out the 2 in then numerator for us, nice !

now our fraction looks like  [2* sin^{2}(A)] / 2 sin(A)cos(A)

so cancel out one of the sines

2*sin(A) / 2 cos(A)

cancel the 2s

Sin(A) / Cos(A) = Tan(A)

nice  it worked out  :P

b)

by the above that we just worked out, then

Tan(15) = Sin(15) / Cos(15)

I had to look up what sin of 15 is b/c it's not one of those special angles but it does have an exact form of

Sin(15) = (√3 - 1) / 2√2

Cos(A) = (√3 + 1) / 2√2

you can use rule of Cos(A-B) = Cos(A)Cos(B)+Sin(A)Sin(B) to get the above and a similar rule for Sin(A-B)

back to our problem,  the 2√2 will cancel out

then we have

Tan(15) =  (√3 - 1) /(√3 + 1)

in the form that is above that's exact, the roots could be approximated but i'll just leave that in the form that is exact.  Most math professors like that form.  

 

4 0
2 years ago
7X(-3)X(-2)^2=?<br> I suck at math
marshall27 [118]

Answer:

-84

Step-by-step explanation:

6 0
3 years ago
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