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Naily [24]
3 years ago
8

Help and show all work please

Mathematics
2 answers:
Anarel [89]3 years ago
7 0

Answer:

1-3

2-8.9

3-6.3

4-26

5-7.8

6-17

7-10

8-32

9-7.9

10-13.2

11- is a right triangle

12- is not a right triangle

13- is a right triangle

14- is not a right triangle

15- is a right triangle

16- is not a right triangle

Step-by-step explanation:

I will do the rest it will just take me some time. I will also give you

step by step at the end.

kari74 [83]3 years ago
3 0

Answer:

1- 3

2- 8.9

3- 6.3

4- 26

5- 7.8

6- 17

7- 10

8- 32

9- 7.9

10- 13.2

11- 289 right triangle because 289 = 289

12- 81 not right triangle

13- 169 right triangle

14- 156 not right triangle

15- 676 right triangle

16- 9 not right triangle

Step-by-step explanation:

a2 + b2 = c2

example:

9x9 + 10x10 = c2

81 + 100 = 181

the square root of 181 = c2

c2 =  

a2 + b2 = c2

a2 = a squared

b2 = b squared

squared means to multiply it by its self

do this for everyone

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If x+z=15x+z=15 and w+y=10w+y=10, what is the value of (3w+3y)(2x+2z)(3w+3y)(2x+2z) ?
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X + z = 15
w + y = 10

(3w + 3y(2x + 2z)  =  3(w + y) * 2(x + z) =  3*10 + 2*15 = 60 Answer
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The amount borrowed by the corporation at 3%. 4% and 5% are $100,000, $250.000 and $150,000 respectively.

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3 years ago
A wheat farmer cuts down the stalks of wheat and gathers them in 200 piles. The 200 gathered piles will be put on a truck. The t
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Answer:

Check the explanation

Step-by-step explanation:

We want to estimate the total weight of grain on the field based on the data on a simple random sample of 5 piles out of 200. The population and sample sizes are N=200 & n=5 respectively.

1) Let Y_1,Y_2,...,Y_{200} be the weight of grain in the 200 piles and y_1,y_2,...,y_{5} be the weights of grain in the pile from the simple random sample.

We know, the sample mean is an unbiased estimator of the population mean. Therefore,

\widehat{\mu}=\overline{y}=\frac{1}{5}\sum_{i=1}^{5}y_i=\frac{1}{5}(3.3+4.1+4.7+5.9+4.5)=4.5

where \mu is the mean weight of grain for all the 200 piles.

Hence, the total grain weight of the population is

\widehat{Y}=Y_1+Y_2+...+Y_{200}

=200\times \widehat{\mu}\: \: \: =200\times 4.5\: \: \: =900\, lbs

2) To calculate a bound on the error of estimates, we need to find the sample standard deviation.

The sample standard deviation is

 S=\sqrt{\frac{1}{5-1}\sum_{i=1}^{5}(y_i-\overline{y})^2}\: \: \: =0.9486

Then, the standard error of \widehat{Y} is

\sigma_{\widehat{Y}} =\sqrt{\frac{N^2S^2}{n}\bigg(\frac{N-n}{N}\bigg)}\: \: \:=83.785

Hence, a 95% bound on the error of estimates is

[\pm z_{0.025}\times \sigma_{\widehat{Y}}]\: \: \: =[\pm 1.96\times 83.875]\: \: \: =[\pm 164.395]

3) Let x_1,x_2,...,x_5 denotes the total weight of the sampled piles.

Mean total weight of the sampled piles is

\overline{x}=\frac{1}{5}\sum_{i=1}^{5}x_i=45

The sample ratio is

r=\frac{\overline{y}}{\overline{x}}=\frac{4.5}{45}=0.1 , this is also the estimate of the population ratio R=\frac{\overline{Y}}{\overline{X}} .

Therefore, the estimated total weight of grain in the population using ratio estimator is

\widehat{Y}_R\: \: =r\times 8800\: \: =0.1\times 8800\: \: =880\, lbs

4) The variance of the ratio estimator is

var(r)=\frac{N-n}{N}\frac{1}{n}\frac{1}{\mu_x^2}\frac{\sum_{i=1}^{5}(y_i-rx_i)^2}{n-1}   , where \mu_x=8800/200=44lbs

=\frac{200-5}{200}\, \frac{1}{5}\: \frac{1}{44^2}\, \frac{0.2}{5-1}=0.000005

Hence, the standard error of the estimate of the total population is

\sigma_R=\sqrt{X^2 \: var(r)}\: \: \: =\sqrt{8800^2\times 0.000005}\: \: \:=21.556

Hence, a 95% bound on the error of estimates is

[\pm z_{0.025}\times \sigma_{R}]\: \: \: =[\pm 1.96\times 21.556]\: \: \: =[\pm 42.25]

8 0
4 years ago
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ASHA 777 [7]

Answer:

16008

Step-by-step explanation:

Sum of an arithmetic sequence is:

S = (n/2) (2a₁ + (n−1) d)

or

S = (n/2) (a₁ + a)

To use either equation, we need to find the number of terms n.  We know the common difference d is 1 − (-9) = 10.  Using the definition of the nth term of an arithmetic sequence:

a = a₁ + (n−1) d

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570 = 10n − 10

580 = 10n

n = 58

Using the first equation to find the sum:

S = (n/2) (2a₁ + (n−1) d)

S = (58/2) (2(-9) + (58−1) 10)

S = 29 (-18 + 570)

S = 16008

Using the second equation to find the sum:

S = (n/2) (a₁ + a)

S = (58/2) (-9 + 561)

S = 16008

4 0
3 years ago
Read 2 more answers
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