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3 years ago

What is equivalent to -2(3x+5)-7x

Mathematics

1 answer:

Blizzard [7]3 years ago

7 0

-13x-10 use the distributive property which allows you to simplify

Hope this helps

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Simplify the following expression: (−4)(−2)(−3) a) −24 B) −11 C) 11 d)24

professor190 [17]

D
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7 0

4 years ago

A golfer keeps track of his score for playing nine holes of golf (half a normal golf round). His mean score is 8080 with a sta

solniwko [45]

Answer:

Therefore the mean and standard deviation of his total score if he plays a full 18 holes are 160 and $11\sqrt2$ respectively.

Step-by-step explanation:

Given that,

For the first 9 holes X:

E(X) = 80

SD(X)=13

For the second 9 holes Y:

E(Y) = 80

SD(Y)=13

For the sum W=X+Y, the following properties holds for means , variance and standard deviation :

E(W)=E(X)+E(Y)

and

V(W)=V(X)+V(Y)

⇒SD²(W)=SD²(X)+SD²(Y) [ Variance = (standard deviation)²]

$\Rightarrow SD(W)=\sqrt{SD^2(X)+SD^2(Y)}$

∴E(W)=E(X)+E(Y) = 80 +80=160

and

∴ $SD(W)=\sqrt{SD^2(X)+SD^2(Y)}$

$=\sqrt{11^2+11^2}$

$=\sqrt{2.11^2}$

$=11\sqrt2$

Therefore the mean and standard deviation of his total score if he plays a full 18 holes are 160 and $11\sqrt2$ respectively.

4 0

4 years ago

Put these numbers in order from least to greatest HELP ASAP TIA

jok3333 [9.3K]

Answer:

bro its so easy

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

18. On Monday, Mr Tan earned a total of $1260 from the sale of some toys. On the next day, he earned a total of $1736 from the s

GalinKa [24]

Answer:

Monday 45

Tuesday 62

Step-by-step explanation:

It's asking you to find the common factor. 28 goes into both of these numbers. So do 1260÷28 and 1736÷28

6 0

3 years ago

Expand the given power using the Binomial Theorem. (10k – m)5

agasfer [191]

Answer:

$(10k - m)^{5}=100000k-50000k^{4}m+10000k^{3}m^{2}-1000k^{2}m^{3}+50km^{4}-m^{5}$

Step-by-step explanation:

* Lets explain how to solve the problem

- The rule of expand the binomial is:

$(a+b)^{n}=(a)^{n}+nC1(a)^{n-1}(b)+nC2(a)^{n-2}(b)^{2}+nC3(a)^{n-3}(b)^{3}+...............+(b)^{5}$

∵ The binomial is $(10k-m)^{5}$

∴ a = 10k , b = -m and n = 5

∴ $(10k-m)^{5}=(10k)^{5}+5C1(10k)^{4}(-m)+5C2(10k)^{3}(-m)^{2}+5C3(10k)^{2}(-m)^{3}+5C4(10k)^{1} (-m)^{4}+5C5(10k)^{0}(-m)^{5}$

∵ 5C1 = 5

∵ 5C2 = 10

∵ 5C3 = 10

∵ 5C4 = 5

∵ 5C5 = 1

∴ $(10k-m)^{5}=100000k^{5}+(5)(10000)k^{4}(-m)+(10)(1000)k^{3}(m^{2})+(10)(100)k^{2}(-m^{3})+5(10k)^{1} (m^{4})+(10k)^{0}(-m^{5})$

∴ $(10k-m)^{5}=100000k^{5}-50000)k^{4}m+10000k^{3}m^{2}-1000k^{2}m^{3}+50km^{4}-m^{5}$

5 0

3 years ago