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2 years ago

The low temperatures for 5 days was 5, -2, -11 ,0 and-3.What was the average low temperature for those days

1 answer:

3 0

Data:
Average (A): ?
Arithmetic mean (A.M) = 5, -2, -11, 0, -3sum of n numbers = 5

Solving:
$A = \frac{AM}{n} = \frac{5+(-2)+(-11)+0+(-3)}{5} = \frac{5-2-11-3}{5} = \frac{\diagup\!\!\!\!3-11-\diagup\!\!\!\!3}{5} = \frac{-11}{5} = - 2.2$

Answer:

<span>The average low temperature of these days is -2.2</span>

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A college parking lot is 140 ft long and 90 ft wide. The college wants to increase the area of the lot by 29% by adding strips o

Novay_Z [31]

The initial dimenssions of the park lot are:

length: 140 ft
width: 90 ft

initial area: 140 * 90 = 12,600 ft^2

Area increased 29% = 12,600 * 1.29 = 16,254 ft^2

width of the strips: x

New length: 140 + x

New width: 90 + x

New area: (140+x)(90+x) = 16,254

Solution of the equation:

12600 + 230x + x^2 = 16254

=> x^2 + 230x - 3654 = 0

Use the quadratic formula.

x = {-230 +/- √[ 230^2 - 4*1*(-3654) ]} / 2 =

x = 14.92

The other solution is negative so it is discarded.

Answer: 15 ft

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2 years ago

I need help bc I keep getting confused

umka2103 [35]

Answer:

6 3/4 (six and three fourths)

Step-by-step explanation:

First, we must turn 4 6/7 into an improper fraction, AKA 28/7. Now, you can't subtract 1/4 from 28/7, can you? So what you need to do is find the greatest common factor of 4 and 7. The GCF would be 28 (7 x 4 = 28, 4 x 7 = 28).

Now we know that the GCF is 28, we can turn the numbers into 28/28 and 1/28. But wait! We aren't done yet. Now we have to change the numerators. Since you multiplied the denominator (7) by 4 to get 28, you must now multiply 28 by 4 as well. (If you do something to the bottom, you have to do it to the top!) Because of this, we are now at 196/28.

Next, we must do the numerator of 1/28. Since we multiplied the denominator (4) by 7 to get 28, we must now multiply 1 by 7. This gets you to 7/28.

It's finally time to subtract. 196 - 7 is 184, so we are at 184/28. But hey, that's an improper fraction! You have to simplify. 184 divided by 28 is 6.75. 0.75 is 3/4 when turned into a fraction.

Mash 6 and 3/4 together, and you've got your answer! 6 3/4.

7 0

2 years ago

120/210 *60 (give the answer in hour and minutes)

Katyanochek1 [597]

120/210 = 0.57
0.57 x 60 = 34.2

34 hours and 12 minutes

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2 years ago

A submarine travels 6.6 km due West from its base and then turns and travels due South for 3.9

geniusboy [140]

Answer:

$7.7$ km

Step-by-step explanation:

The submarine's path from its base forms a right triangle when its final position is "connected" to the base. We know that the right triangle has legs of $6.6$ km and $3.9$ km, and we need to find the length of its hypotenuse. To do so, we can use the Pythagorean Theorem, which states that in a right triangle, $a^{2}+b^{2} =c^{2}$ , where $a$ and $b$ are the lengths of the triangle's legs and $c$ is the length of the triangle's hypotenuse. In this case, we know what $a$ and $b$ are, and we need to solve for c, so after substituting the given values of $a$ and $b$ into $a^{2}+b^{2} =c^{2}$ to solve for c, we get:

$a^{2}+b^{2} =c^{2}$

$6.6^{2} +3.9^{2} =c^{2}$ (Substitute $a=6.6$ and $b=3.9$ into the equation)

$43.56+15.21=c^{2}$ (Evaluate the squares on the LHS)

$58.77=c^{2}$ (Simplify the LHS)

$c^{2}=58.77$ (Symmetric Property of Equality)

$\sqrt{c^{2} } =\sqrt{58.77}$ (Take the square root of both sides of the equation)

$c=7.7,c=-7.7$ (Simplify)

$c=-7.7$ is an extraneous solution because you can't have negative distance, if that makes sense, so therefore, the submarine is approximately $7.7$ km away from its base. Hope this helps!