The equation of the line perpendicular to 5x+2y=20 and containing the point (5,10) is:

Can someone help me and explain this Algebra 2?

Tju [1.3M]

Nth term of gp = a(1 - r^n)/(1 - r), where n∞ = 10, n = ∞, a = 2, r < 1
10 = 2(1 - r^∞)/(1 - r)
10 = 2/(1 - r)
1 - r = 2/10 = 1/5
r = 1 - 1/5 = 4/5
An infinite deometric series with a beginning value of 2 that converges to 10 has common ratio of 4/5.

The first four terms are: 2, 1.6, 1.28, 1.024

5 0

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

denpristay [2]

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

Read 2 more answers

I don’t understand 12 can someone help me

slamgirl [31]

Pretty sure its D . the amount he charges for his service and hours must be less than or equal to 225

4 0

3 years ago

Read 2 more answers

The National Highway Traffic Safety Administration reported the percentage of traffic accidents occurring each day of the week.

Zigmanuir [339]

Answer:

Hence,

a)

Test statistic $X^{2}$ = 9.269.

p value = 0.148

b)

Sunday= 15.23%, Monday= 12.62%, Tuesday= 13.10%, Wednesday=11.67%,

Thursday= 13.10%, Friday= 16.19%, Saturday= 18.09%.

Step-by-step explanation:

a)

relative observed Expected Chi square

Category frequency(p) $O_{i}$ Ei=total*p $(O-E)^{2}$ $R^{2}_{i} =(O_{i} -E_{i} )^{2} /E_{i}$

1 0.142857143 64.0 60.00 16.00 0.267

2 0.142857143 53.0 60.00 49.00 0.817

3 0.142857143 55.0 60.00 25.00 0.417

4 0.142857143 49.0 60.00 121.00 2.017

5 0.142857143 55.0 60.00 25.00 0.417

6 0.142857143 68.0 60.00 64.00 1.067

7 0.142857143 76.0 60.00 256.00 4.267

Total 1.000 420 420 556 9.269

Test statistic $X^{2}$ = 9.269.

p value = 0.148 from excel: chidist(9.269,6)

b)

Percentage %= Frequency of traffic occur on category x 100 / Total number of frequency given.

Category Percentage%

Sunday 15.23%

Monday 12.62%

Tuesday 13.10%

Wednesday 11.67%

Thursday 13.10%

Friday 16.19%

Saturday 18.09%

5 0

3 years ago

Name the values of the given digits the 4s in a 4400

IRISSAK [1]

The values of 4s that are present in 4400 are 4 thousands and 4 hundreds.
4400 = 4 thousands + 4 hundreds + 0 tens + 0 ones
so the value of first 4 is 4 thousands and second 4 is 4 hundreds.

5 0

4 years ago