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3 years ago

7s. 49

Mathematics

1 answer:

Greeley [361]3 years ago

4 0

The correct question is
(7s)/(s^2 - 14s + 49) - (49)/(s^2 - 14s + 49)

----- > (7s-49)/(s^2 - 14s + 49)

s^2 - 14s + 49------------ > <span>solving the quadratic equation
</span>
s1=7 s2=7 ------------> see the attached figure
therefore
s^2 - 14s + 49=(s-7)(s-7)=(s-7)²
<span>substituting
</span>(7s-49)/(s^2 - 14s + 49)=(7s-49)/(s-7)²=7[s-7]/[(s-7)²]------ > 1/(s-7)

the answer is 1/(s-7)

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John has a rectangular-shaped field whose length is 62.5 yards and width is 45.3 yards. The area of the field is ________ square

oee [108]

Area = 2831.25 square yards

Perimeter =215.6 yards

EXPLANATION

The area and perimeter of a rectangular field are found using the formula for finding the area and perimeter of a rectangle respectively.

That means, area of the rectangular field is given by the formula,

$A= l\times w$

We just have to substitute
$l=62.5$ and $w= 45.3$ into the given formula and evaluate.

This implies that;

$A= 62.5\times 45.3$

This gives the area of the rectangular-shaped field to be;

$A= 2831.25$ square yards.

Now for the perimeter, we use the formula

$P=2w +2l$

Or

$P=2(w +l)$

Substituting the values for the length and width gives,

$P=2(62.5+45.3)$

$\Rightarrow P=2(107.8)$

$\Rightarrow P=215.6$

Hence the perimeter of the rectangular shaped field is 215.6 yards.

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3 years ago

Subtract 5x^2-2x-55x 2 −2x−5 from -2x^2-5x-7−2x 2 −5x−7.

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Answer:

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Step-by-step explanation:

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3 years ago

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lys-0071 [83]

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he amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.3 minutes and s

sp2606 [1]

Complete question:

He amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.3 minutes and standard deviation 1.4 minutes. Suppose that a random sample of n equals 47 customers is observed. Find the probability that the average time waiting in line for these customers is

a) less than 8 minutes

b) between 8 and 9 minutes

c) less than 7.5 minutes

Answer:

a) 0.0708

b) 0.9291

c) 0.0000

Step-by-step explanation:

Given:

n = 47

u = 8.3 mins

s.d = 1.4 mins

a) Less than 8 minutes:

$P(X$

P(X' < 8) = P(Z< - 1.47)

Using the normal distribution table:

NORMSDIST(-1.47)

= 0.0708

b) between 8 and 9 minutes:

P(8< X' <9) = $[\frac{8-8.3}{1.4/ \sqrt{47}}< \frac{X'-u}{s.d/ \sqrt{n}} < \frac{9-8.3}{1.4/ \sqrt{47}}]$

= P(-1.47 <Z< 6.366)

= P( Z< 6.366) - P(Z< -1.47)

Using normal distribution table,

$NORMSDIST(6.366)-NORMSDIST(-1.47)$

0.9999 - 0.0708

= 0.9291

c) Less than 7.5 minutes:

P(X'<7.5) = $P [Z< \frac{7.5-8.3}{1.4/ \sqrt{47}}]$

P(X' < 7.5) = P(Z< -3.92)

NORMSDIST (-3.92)

= 0.0000

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3 years ago

2x+3y=41 and 5x+4y=85

Sever21 [200]

Answer:

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Slope = -2.500/2.000 = -1.250

x-intercept = 85/5 = 17

y-intercept = 85/4 = 21.25000

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Slope = -1.333/2.000 = -0.667

x-intercept = 41/2 = 20.50000

y-intercept = 41/3 = 13.66667

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3 years ago