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a_sh-v [17]
4 years ago
12

I need help can you please help me out?

Mathematics
1 answer:
zalisa [80]4 years ago
7 0
The lcm of 3,4,5 is 60. heres a trick if it helps u can multiply 3*5*4 
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8-4x=4-3(2x+6) ??? Someone help
Maurinko [17]
<span>       8 - 4x=4 - 3(2x+6)
<=> 8 - 4x = 4 - 6x - 18
<=> 8 - 4 + 18 = 4x - 6x
<=> 22 = -2x
x = -11</span>
7 0
3 years ago
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An article claims that 12% of trees are infested by a bark beetle. A random sample of 1,000 trees were tested for traces of the
inna [77]

Answer: 0.6812

Step-by-step explanation:

Let p be the population proportion of trees are infested by a bark beetle.

As per given: p= 12%= 0.12

Sample size : n= 1000

Number of trees affected in sample = 1000

Sample proportion of trees are infested by a bark beetle. = \hat{p}=\dfrac{127}{1000}=0.127

Now, the z-test statistic : z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}

So, z=\dfrac{0.127-0.12}{\sqrt{\dfrac{0.12\times 0.88}{1000}}}

z=\dfrac{0.007}{\sqrt{\dfrac{0.1056}{1000}}}\\\\\\=\dfrac{0.007}{\sqrt{0.0001056}}\\\\\\=\dfrac{0.007}{0.010276}\approx0.6812

Hence, the value of the z-test statistic = 0.6812 .

7 0
4 years ago
Subtract 5x^2-2x-55x 2 −2x−5 from -2x^2-5x-7−2x 2 −5x−7.
leva [86]

Answer:

7x^2 -128x -19

Step-by-step explanation:

5 0
3 years ago
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Does anyone know how to get the answer for this?
klasskru [66]
B!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

K K K K K
K K K K
7 0
3 years ago
How to work it out logically? Question a !!!
Arturiano [62]

Answer:

Fencing is done along KL which is (1500+520.8=2020.8 m) from the top left corner and divides the property into half.

Step-by-step explanation:

Given the figure with dimensions. we have to find the area of given figure.

Area of figure=ar(1)+ar(2)+ar(3)

Area of region 1 = ar(ANGI)+ar(AIB)

                          =L\times B+\frac{1}{2}\times base\times height\\\\=[1500\times (5000-2000-1500)]+\frac{1}{2}\times (3000-1500)\times (5000-2000-1500)\\\\=3375000m^2=337.5ha

Area of region 2 = ar(DHBC)

                       =2000\times1500\\\\=3000000m^2=300ha

Area of region 3 = ar(GFEH)

                             (2000+1500)\times 1000\\\\=3500000m^2=350ha

Hence, Area of figure=ar(1)+ar(2)+ar(3)=337.5ha+300ha+350ha

                                                 =987.5 ha

Now, we have to do straight-line fencing such that area become half and cost of fencing is minimum.

Let the fencing be done through x m downward from B which divides the two into equal area.

⇒ Area of upper part above fencing=Area of lower part below fencing

⇒ar(ANGB)+ar(GKLB)=ar(KLCM)+ar(MDCF)\\\\337500+3000x=(3500-x)\times 1000+2000(1500-x)\\\\3375000+3000x=3500000-1000x+3000000-2000x\\\\6000x=315000\\\\x=\frac{315000}{6000}=520.8m

Hence, fencing is done along KL which is (1500+520.8=2020.8 m) from the top left corner and divides the property into half.

7 0
3 years ago
Read 2 more answers
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