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Mnenie [13.5K]
2 years ago
7

I will pay to find the answer

Mathematics
1 answer:
fenix001 [56]2 years ago
5 0

Step-by-step explanation:

9) C

10) B

11) (b-5)(b+8)

12) (7w - 2)²

13) 14(5x + 2)(3x - 1)

14) (4h - 1) and (h + 6)

15) 3 and (a - 2) and (a + 8)

16) 5 and (4w - 1) and (w + 2)

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How would you solve -7x-4y=16
Anni [7]
So one easy way is to make the equation equal to only one place holder so 
-7x-4y=16
add 7x to both sdies
-4y=16+7x
multiply both sides by -1
4y=-16-7x
divide both sides by 4
y=-4-7/4x
subsitute values for x and get values for y
exg
if x=0 then y=-4
if x=1 then y=-5 3/7
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8 0
3 years ago
What is the period of the function y = sin(4x) - 3?
iren [92.7K]

Answer:

\frac{\pi }{2}

3 0
3 years ago
Two surfers and statistics students collected data on the number of days on which surfers surfed in the last month for 30 longbo
Alisiya [41]

Answer:

Do not reject H0. The mean days surfed for longboarders is significantly larger than the mean days surfed for all shortboarders

Step-by-step explanation:

The null hypothesis is that  the mean days surfed for all long boarders is larger than the mean days surfed for all short boarders

H0:  μL > μs      against the claim Ha:  μL≤ μs

the alternate hypothesis is the mean days surfed for all long boarders isless or equal to  the mean days surfed for all short boarders (because long boards can go out in many different surfing conditions)

The test statistic is

t= x1- x2/  √s1/n1+ s2/n2

1) Calculations

Longboards

Mean

ˉx=∑x/n=4+8+9+4+9+7+9+6+6+11+15+13+16+12+10+12+18+20+15+10+15+19+21+9+22+19+23+13+12+10/30

=377/30

=12.5667

Longboard Variance S2=[∑dx²-(∑dx)²/n]/n-1

=[831-(-13)²/30]/29

=831-5.6333/29

=825.3667/29

=28.4609

Shortboard Mean

ˉx=∑x/n=6+4+6+6+7+7+7+10+4+6+7+5+8+9+4+15+13+9+12+11+12+13+9+11+13+15+9+19+20+11/30

=288/30

=9.6

Shortboard Variance S2=[∑x²-(∑x)²/n]/n-1

=[ 3270-(288)2/30]/29

=3270-2764.8/29

=505.2/29

=17.4207

2) Putting values in the test statistic

t=|x1-x2|/√S²1/n1+S²2/n2

t =|12.5667-9.6|/√28.4609/30+17.4207/30

t =|2.9667|/√0.9487+0.5807

t=|2.9667|/√1.5294

t=|2.9667|/1.2367

t=2.3989

3) Degree of freedom =n1+n2-2=30+30-2=58

4) The critical region is t ≤ t(0.05) (58) =1.6716

5) Since the calculated t= 2.4 does not fall in the critical region t(0.05) (58)  ≤ 1.6716 we do not reject H0.

The p-value is 0.008969. The result is significant at p <0 .05.

6 0
3 years ago
What is the missing constant term in the perfect square that starts with x2+18xx^2+18xx2+18xx, squared, plus, 18, x ?
katrin2010 [14]

Answer:

81

Step-by-step explanation:

Given the expression: x^2+18x

To make it a perfect square, we follow the following steps.

Divide the coefficient of x by 2

  • Coefficient of x=18, 18/2=9

Square It

  • 9 X 9=81

Add it to the expression

  • x^2+18x+81

This is written as a perfect square as: (x+9)^2

The constant term required therefore is 81.

4 0
3 years ago
Read 2 more answers
Let T : V → W be a linear transformation from vector space V into vector space W.
wel

Explanation:

Since {v1,...,vp} is linearly dependent, there exist scalars a1,...,ap, with not all of them being 0 such that a1v1+a2v2+...+apvp = 0. Using the linearity of T we have that

a1*T(v1)+a2*T(v1) + ... + ap*T(vp) = T(a1v19+T(a2v2)+...+T(avp) = T(a1v1+a2v2+...+apvp) = T(0) = 0.

Since at least one ai is different from 0, we obtain a non trivial linear combination that eliminates T(v1) , ..., T(vp). That proves that {T(v1) , ..., T(vp)} is a linearly dependent set of W.

7 0
3 years ago
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