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blagie [28]
3 years ago
6

The mean value of land and buildings per acre from a sample of farms is ​$1500​, with a standard deviation of ​$200. the data se

t has a​ bell-shaped distribution. assume the number of farms in the sample is 72. ​(a) use the empirical rule to estimate the number of farms whose land and building values per acre are between ​$1300 and ​$1700. nothing farms ​(round to the nearest whole number as​ needed.) ​(b) if 22 additional farms were​ sampled, about how many of these additional farms would you expect to have land and building values between ​$1300 per acre and ​$1700 per​ acre? nothing farms out of 22 ​(round to the nearest whole number as​ needed.)
Mathematics
1 answer:
MArishka [77]3 years ago
6 0
I need help! someone answer this problem asap
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I really need help here... please
Vikentia [17]

XW must be congruent to YZ, and WZ must be congruent to XY

In order for it to be a parallelogram, the opposite sides must be congruent.

8 0
3 years ago
What's (7x/2-5x+3)+(2x/2+4x-6)
Anika [276]
I'm assuming that when you wrote "(7x/2-5x+3)+(2x/2+4x-6)," you actually meant "<span>(7x^2-5x+3)+(2x^2+4x-6).  Correct me if I'm wrong here.

</span><span>+(7x^2-5x+3)
</span><span>+(2x/2+4x-6)
-------------------
=9x^2 - x - 3 (answer) </span>
5 0
3 years ago
The formula K=59(F−32)+273.15 converts temperatures from Fahrenheit F to Kelvin K.
Semmy [17]

Answer:

9/5 (K-273.15) + 32=F

Step-by-step explanation:

K=5 /9(F−32)+273.15

Subtract 273.15 from each side

K-273.15=5/9(F−32)+273.15-273.15

K-273.15=5/9(F−32)

Multiply by 9/5 on each side

9/5 (K-273.15)= 9/5 *5/9(F−32)

9/5 (K-273.15)=(F−32)

Add 32 to each side

9/5 (K-273.15) + 32=F−32 +32

9/5 (K-273.15) + 32=F

4 0
3 years ago
Read 2 more answers
PLEASE HELP ME! WILL MARK BRAINlIEST!!
vivado [14]
Answer:

1.) Increase

2.) Decrease

3.) Increase

4.) Decrease
6 0
3 years ago
Media experts claim that daily print newspapers are declining because of Internet access. Listed​ below, from left to right and
yKpoI14uk [10]

Answer:

(a) The median is 1478

(b) Null hypothesis H₀ : μ₁ = μ₂

Alternative hypothesis Hₐ : μ₁ > μ₂

(c) The test statistic  t_{\alpha/2} is 6.678155

(d) The p value is  1.2×10⁻¹¹

(e) We reject our null hypothesis H₀. In terms of the test, there is sufficient evidence to suggest that there is a trend in the numbers of daily newspaper because the probability for a decrease in the numbers is very high

Step-by-step explanation:

(a) Here we have the data as follows;

1623 1586 1574 1554 1544 1531 1519 1513 1491 1484 1478 1471 1458 1456 1458 1453 1438 1428 1405 1395 1377

The median is = 1478

Therefore we have above the median  

1623 1586 1574 1554 1544 1531 1519 1513 1491 1484

The mean,  \bar{x}_{1}= 1541.9

Standard deviation, σ₁ = 41.38224257

n₁ = 10

Below the median  

1471 1458 1456 1458 1453 1438 1428 1405 1395 1377

The mean,  \bar{x}_{2}}= 1433.9

Standard deviation, σ₂ = 30.04812806

n₂ = 10

(b) Null hypothesis H₀ : μ₁ = μ₂

Alternative hypothesis Hₐ : μ₁ > μ₂

(c) The formula for t test is given by;

t_{\alpha/2} =\frac{(\bar{x}_{1}-\bar{x}_{2})}{\sqrt{\frac{\sigma_{1}^{2} }{n_{1}}-\frac{\sigma _{2}^{2}}{n_{2}}}}

df = 10 - 1 = 9, α = 0.05

Therefore, the test statistic  t_{\alpha/2} = 6.678155

(d) The p value from statistical relations is Probability p = 1.2×10⁻¹¹

Critical z at 5% confidence level = 1.645

Since P << 0.05, e reject the

(e) Therefore, we reject our null hypothesis H₀. In terms of the test, there is sufficient evidence to suggest that there is a trend in the numbers of daily newspaper because the probability for a decrease in the numbers is very high.

5 0
3 years ago
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