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2 years ago

5

A square rug covers 79 square feet of floor. What is the approximate length of one side of the rug? (Approximate to the nearest

hundredth foot.)

1 answer:

Mice21 [21]2 years ago

8 0

A = a^2.....where " a " is the length of one side
A = 79

79 = a^2...take sqrt of both sides, eliminating the ^2
sqrt 79 = a
8.89 = a

so the length of one side is approximately 8.89 ft <==

You might be interested in

Lucy wants to have a fun-filled day without spending too much money. The local zoo charges a $25 entry fee and a parking fee of

ehidna [41]

Answer:

The required system of equations is,

$y=25+x$

$y=20+2x$ .

Step-by-step explanation:

Let the number of hours spent by Lucy = x

It is given that,

Local zoo charges initial fee $25 and per hour fee $1.

So, the equation representing the cost if Lucy stays 'x' hours is,

$y=25+x$ in dollars.

Also, Local citcus charges initial fee $20 and per hour fee $2.

So, the equation representing the cost if Lucy stays 'x' hours is,

$y=20+2x$ in dollars.

Since, the objective function for the situation is to 'minimize the cost'.

So, upon solving the equations, we will get the minimum value as,

25+x = 20+2x

i.e. x= 5

Thus,

$y=25+x$ implies $y=25+5$ i.e. y= $30

$y=20+2x$ implies $y=20+2\times 5$ i.e. $y=20+10$ i.e. y= $30

Thus, the minimum value is $30.

The required system of equations is,

$y=25+x$

$y=20+2x$ .

8 0

2 years ago

4. Hydrocarbons in the cab of an automobile were measured during trips on the New Jersey Turnpike and trips through the Lincoln

pochemuha

Answer:

No these these result do not differ at 95% confidence level

Step-by-step explanation:

From the question we are told that

The first concentrations is $c _1= 30.0 \ g/m^3$

The second concentrations is $c _2 = 52.9 \ g/m^3$

The first sample size is $n_1 = 32$

The second sample size is $n_2 = 32$

The first standard deviation is $\sigma_1 = 30.0$

The first standard deviation is $\sigma_1 = 29.0$

The mean for Turnpike is $\= x _1 = \frac{c_1}{n} = \frac{31.4}{32} = 0.98125$

The mean for Tunnel is $\= x _2 = \frac{c_2}{n} = \frac{52.9}{32} = 1.6531$

The null hypothesis is $H_o : \mu _1 - \mu_2 = 0$

The alternative hypothesis is $H_a : \mu _1 - \mu_2 \ne 0$

Generally the test statistics is mathematically represented as

$t = \frac{\= x_1 - \= x_2}{ \sqrt{\frac{\sigma_1^2}{n_1} +\frac{\sigma_2^2}{n_2} }}$

$t = \frac{0.98125 - 1.6531}{ \sqrt{\frac{30^2}{32} +\frac{29^2}{32} }}$

$t = - 0.0899$

Generally the degree of freedom is mathematically represented as

$df = 32+ 32 - 2$

$df = 62$

The significance $\alpha$ is evaluated as

$\alpha = (C - 100 )\%$

=> $\alpha = (95 - 100 )\%$

=> $\alpha =0.05$

The critical value is evaluated as

$t_c = 2 * t_{0.05 , 62}$

From the student t- distribution table

$t_{0.05, 62} = 1.67$

So

$t_c = 2 * 1.67$

=> $t_c = 3.34$

given that

$t_c > t$ we fail to reject the null hypothesis so this mean that the result do not differ

6 0

2 years ago

The flathead rail tunnel in Montana is about 7 3/4 miles long. A train travels at a speed of 3/4 mile per mile minute.how long w

kogti [31]

7 3/4 × 3/4 = So u add 3×3=9
Then add 4×4=16 Give you 9/16
Then you add that 7 you left out with 9/16
But just the 9 not the 16
7 × 9 = 63 And your answer is 63/16
Hope i help :-)

5 0

2 years ago

Read 2 more answers

Arrange these functions from the greatest to the least value based on the average rate of change in the specified interval.Tiles

Ugo [173]

By definition, the average rate of change is given by:

$AVR = \frac{f(x2)-f(x1)}{x2-x1}$

We evaluate each of the functions in the given interval.

We have then:

For f (x) = x ^ 2 + 3x:

Evaluating for x = -2:

$f (-2) = (-2) ^ 2 + 3 (-2)\\f (-2) = 4 - 6\\f (-2) = - 2$

Evaluating for x = 3:

$f (3) = (3) ^ 2 + 3 (3)\\f (3) = 9 + 9\\f (3) = 18$

Then, the AVR is:

$AVR = \frac{18-(-2)}{3-(-2)}$

$AVR = \frac{18+2}{3+2}$

$AVR = \frac{20}{5}$

$AVR = 4$

For f (x) = 3x - 8:

Evaluating for x =4:

$f (4) = 3 (4) - 8\\f (4) = 12 - 8\\f (4) = 4$

Evaluating for x = 5:

$f (5) = 3 (5) - 8\\f (5) = 15 - 8\\f (5) = 7$

Then, the AVR is:

$AVR = \frac{7-4}{5-4}$

$AVR = \frac{3}{1}$

$AVR = 3$

For f (x) = x ^ 2 - 2x:

Evaluating for x = -3:

$f (-3) = (-3) ^ 2 - 2 (-3)\\f (-3) = 9 + 6\\f (-3) = 15$

Evaluating for x = 4:

$f (4) = (4) ^ 2 - 2 (4)\\f (4) = 16 - 8\\f (4) = 8$

Then, the AVR is:

$AVR = \frac{8-15}{4-(-3)}$

$AVR = \frac{-7}{4+3}$

$AVR = \frac{-7}{7}$

$AVR = -1$

For f (x) = x ^ 2 - 5:

Evaluating for x = -1:

$f (-1) = (-1) ^ 2 - 5\\f (-1) = 1 - 5\\f (-1) = - 4$

Evaluating for x = 1:

$f (1) = (1) ^ 2 - 5\\f (1) = 1 - 5\\f (1) = - 4$

Then, the AVR is:

$AVR = \frac{-4-(-4)}{1-(-1)}$

$AVR = \frac{-4+4}{1+1}$

$AVR = \frac{0}{2}$

$AVR = 0$

Answer:

from the greatest to the least value based on the average rate of change in the specified interval:

f(x) = x^2 + 3x interval: [-2, 3]

f(x) = 3x - 8 interval: [4, 5]

f(x) = x^2 - 5 interval: [-1, 1]

f(x) = x^2 - 2x interval: [-3, 4]

4 0

2 years ago

Mateus’s bank issued an advertisement saying that 90\%90%90, percent of its customers are satisfied with the bank’s services. Si

densk [106]

Answer:

The probability of getting a sample with 80% satisfied customers or less is 0.0125.

Step-by-step explanation:

We are given that the results of 1000 simulations, each simulating a sample of 80 customers, assuming there are 90 percent satisfied customers.

Let $\hat p$ = <u><em>sample proportion of satisfied customers</em></u>

The z-score probability distribution for the sample proportion is given by;

Z = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, p = population proportion of satisfied customers = 90%

n = sample of customers = 80

Now, the probability of getting a sample with 80% satisfied customers or less is given by = P( $\hat p \leq$ 80%)

P( $\hat p \leq$ 80%) = P( $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ $\leq$ $\frac{0.80-0.90}{\sqrt{\frac{0.80(1-0.80)}{80} } }$ ) = P(Z $\leq$ -2.24) = 1 - P(Z < 2.24)

= 1 - 0.9875 = <u>0.0125</u>

The above probability is calculated by looking at the value of x = 2.24 in the z table which has an area of 0.9875.

8 0

2 years ago