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Soloha48 [4]
2 years ago
14

The diagram below represents the measurements of Jennie’s yard. The yard’s width is x feet shorter than its length. If the area

of the yard is 540 square feet, how many feet shorter than the length is the width?
A yard with a length of 30 feet and a width of (30 minus x) feet.

Which equation represents the scenario?
30 + (30 – x) = 540
540 – 30 = 30 – x
30(30 – x) = 540
540 – (30 – x) = 30

Mathematics
2 answers:
larisa [96]2 years ago
5 0

Answer:

30(30-x)=540

Step-by-step explanation:

AleksandrR [38]2 years ago
3 0

Answer:

30(30 – x) = 540

c

Step-by-step explanation:

hi , which grade r u in ?

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Graphs of the following equations are straight lines except :
scoundrel [369]

Answer:

D

Step-by-step explanation:

6 0
2 years ago
A liquid vessel which is initially empty is in the form of an inverted regular hexagonal pyramid of altitude 25 feet and base ed
Misha Larkins [42]
<span>Since: v =sqrt(3)/2 s^2h 

6779 liters x 0.0353cu ft/1 liter= 239.299 cu ft 
but by proportion s/h = 10/25 
s = 10/25 h 
and v = sqrt(3)/2 (10/25 h)^2 h 
239.299 = 0.139 h^3 

h = (239.299/0.139)^(1/3) = 11.985 ft</span>
6 0
2 years ago
Read 2 more answers
The extract of a plant native to Taiwan has been tested as a possible treatment for Leukemia. One of the chemical compounds prod
True [87]

Answer:

a) 57.35%

b) 99.99%

c) 68.27%

Step-by-step explanation:

When we have a random variable X that is normally distributed with mean \large\bf \mu and standard deviation \large\bf \sigma, then  

The probability that the random variable has a value less than a, P(X < a) = P(X ≤ a) is the area under the normal curve with mean \large\bf \mu and standard deviation \large\bf \sigma to the left of a.

The probability that the random variable has a value greater than b, P(X > b) = P(X ≥ b) is the area under the normal curve with mean \large\bf \mu and standard deviation \large\bf \sigma to the right of b.

The probability that the random variable has a value between a and b, P(a < X < b) = P(a ≤ X ≤  b) = P(a < X ≤  b)= P(a ≤ X < b) is the area under the normal curve with mean \large\bf \mu and standard deviation \large\bf \sigma between a and b.

In this case, the random variable is the collagen amount found in the extract of the plant. The mean is 63 g/ml and the standard deviation is 5.4 g/ml

(a) What is the probability that the amount of collagen is greater than 62 grams per mililiter?

As we have seen, we need to find the area under the normal curve with mean 63 and standard deviation 5.4 to the right of 62 (see picture).

You can find this value easily with a calculator or a spreadsheet. If you prefer the old-style, then you have to standardize the values and look up in a table.

<em>If you have access to Excel or OpenOffice Calc, you can find this value by introducing the formula: </em>

<em>1- NORMDIST(62,63,5.4,1) in Excel </em>

<em>1 - NORMDIST(62;63;5.4;1) in OpenOffice Calc </em>

<em>and we will get a value of 0.5735 or 57.35% </em>

(b) What is the probability that the amount of collagen is less than 90 grams per mililiter?

Now we want the area to the left of 90

<em>NORMDIST(90,63,5.4,1) in Excel </em>

<em>NORMDIST(90;63;5.4;1) in OpenOffice Calc </em>

You will get a value of 0.9999 or 99.99%

(c) What percentage of compounds formed from the extract of this plant fall within 1 standard deviations of the mean?

You can use either the rule that 68.27% of the data falls between \large\bf \mu -\sigma and \large\bf \mu +\sigma or compute area between 63 - 5.4 and 63 + 5.4, that is to say, the area between 57.6 and 68.4  

<em>In Excel </em>

<em>NORMDIST(68.4,63,5.4,1) - NORMDIST(57.6,63,5.4,1)  </em>

<em>In OpenOffice Calc  </em>

<em>NORMDIST(68.4;63;5.4;1) - NORMDIST(57.6;63;5.4;1)  </em>

In any case we get a value of 0.6827 or 68.27%

3 0
2 years ago
Don't understand this Please help asap
madam [21]
This is easy what do you need help with?
8 0
2 years ago
Lucia knows the fourth term in a sequence is 55 and the ninth term in the same sequence is 90. Explain how she can find the comm
Andreyy89
Using the formula for the nth term of an arithmetic progression.
an = a + (n - 1)d
a(4) = a + 3d = 55
a(9) = a + 8d = 90
a(9) - a(4) => 5d = 35
d = 35/5 = 7.
From a(4): a = 55 - 3d = 55 - 3(7) = 55 - 21 = 34

a(2) = a + d = 34 + 7 = 41.

4 0
2 years ago
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