Can any one help with this math problem: −1/2−(−5/9) I am a little confused

The length of a rectangular floor is 4 feet longer than its width w. The area of the floor is 525 ft^2. A) Write a quadratic equ

mina [271]

Answer:

x^21x+25x-525-0

x^21x+25x-525-0xx^2 - 3.7_) +5^2

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-21

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0(x-21)(x+25)=0

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0(x-21)(x+25)=0X-21=0

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0(x-21)(x+25)=0X-21=0x+25= 0

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0(x-21)(x+25)=0X-21=0x+25= 0x-21+21=21

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0(x-21)(x+25)=0X-21=0x+25= 0x-21+21=21(x+25)-25=-25

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0(x-21)(x+25)=0X-21=0x+25= 0x-21+21=21(x+25)-25=-25x=21

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0(x-21)(x+25)=0X-21=0x+25= 0x-21+21=21(x+25)-25=-25x=21x+25-25=-25

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0(x-21)(x+25)=0X-21=0x+25= 0x-21+21=21(x+25)-25=-25x=21x+25-25=-25x= 21

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0(x-21)(x+25)=0X-21=0x+25= 0x-21+21=21(x+25)-25=-25x=21x+25-25=-25x= 21x= -25

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0(x-21)(x+25)=0X-21=0x+25= 0x-21+21=21(x+25)-25=-25x=21x+25-25=-25x= 21x= -25ensiah193

4 0

2 years ago

A data set of 5 observations for Concession Sales per person at a theater and Minutes before the movie begins results in the est

otez555 [7]

Answer:

Check the explanation

Step-by-step explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

5 0

3 years ago

The accompanying summary data on total cholesterol level (mmol/l) was obtained from a sample of Asian postmenopausal women who w

dimulka [17.4K]

Answer:

99% CI:

$-4.637\leq\mu_v-\mu_o\leq 3.737$

Step-by-step explanation:

We have to calculate a 99%CI for the difference of means for the vegan and the omnivore.

First, we have to estimate the standard deviation

$s_{Md}=\sqrt{\frac{2MSE}{n_h}}$

The MSE can be calculated as

$MSE=\frac{(SSE_1+SSE_2)}{df} =\frac{(n_1s_1)^2+(n_2*s_2)^2}{n_1+n_2-2}\\\\MSE=\frac{(85*1.05)^2+(96*1.20)^2}{85+96-2}=\frac{7965.56+13272.04}{179} =118.65$

The weighted sample size nh can be calculated as

$n_h=\frac{2}{1/n_1+1/n_2}=\frac{2}{1/85+1/96}=\frac{2}{0.0222} =90.16$

The standard deviation then becomes

$s_{Md}=\sqrt{\frac{2MSE}{n_h}}=\sqrt{\frac{2*118.65}{90.16}}=\sqrt{2.632} =1.623$

The z-value for a 99% confidence interval is z=2.58.

The difference between means is

$\Delta M=M_v-M_o=5.10-5.55=-0.45$

Then the confidence interval can be constructed as

$\Delta M-z*s_{Md}\leq\mu_v-\mu_o\leq \Delta M+z*s_{Md}\\\\ -0.45-2.58*1.623\leq\mu_v-\mu_o\leq -0.450+2.58*1.623\\\\-0.450-4.187\leq\mu_v-\mu_o\leq-0.450+4.187\\\\-4.637\leq\mu_v-\mu_o\leq 3.737$