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neonofarm [45]
2 years ago
13

The "normal" high temperature for a June day in Albemarle is $82^\circ$ (degrees Fahrenheit). Last year, the first ten days of J

une were unusually hot, with highs of $84^\circ,$ $88^\circ,$ $91^\circ,$ $92^\circ,$ $96^\circ,$ $94^\circ,$ $89^\circ,$ $78^\circ,$ $87^\circ,$ and $81^\circ$. How many degrees above normal was the average high for that 10-day period?
Mathematics
1 answer:
crimeas [40]2 years ago
8 0

Answer:

6° F

Step-by-step explanation:

To get average of 10 days highs, we need to add all and divide by 10.

Average = \frac{84+88+91+92+96+94+89+78+87+81}{10}\\=\frac{880}{10}\\=88

And, the normal high temp is given as 82.

<em>How much higher is 88 than 82??</em>

<em />

It is 88 - 82 = 6°F above

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starting on the positive x axis and moving 1/3 of the way around the circle in 3 m radius, ow far will it go?
nlexa [21]
You will move 1/3 of the circumference of the circle with a radius of 3m.

C=2πr  so C/3 is:

d=2πr/3, we are told that the radius is 3m so:

d=2π3/3

d=2π m

d≈6.28 m (to nearest hundredth of a meter)
7 0
2 years ago
If h:k = 2:5, x:y=3:4 and 2h+x:k+2y = 1:2, find the ratio h-x: k-y.​
fenix001 [56]

Given:

h:k=2:5,x:y=3:4,2h+x:k+2y=1:2

To find:

h-x:k-y

Solution:

We have, h:k=2:5 and x:y=3:4.

Let the values of h and k are 2a and 5a respectively.

Let the values of x and y are 3b and 4b respectively.

We have,

2h+x:k+2y=1:2

It can be written as

\dfrac{2h+x}{k+2y}=\dfrac{1}{2}

\dfrac{2(2a)+(3b)}{5a+2(4b)}=\dfrac{1}{2}

\dfrac{4a+3b}{5a+8b}=\dfrac{1}{2}

2(4a+3b)=1(5a+8b)

On further simplification, we get

8a+6b=5a+8b

8a-5a=8b-6b

3a=2b

a=\dfrac{2b}{3}

Now,

h-x:k-y=\dfrac{h-x}{k-y}

h-x:k-y=\dfrac{2-3b}{5a-4b}

h-x:k-y=\dfrac{2\times \dfrac{2b}{3}-3b}{5\times \dfrac{2b}{3}-4b}

h-x:k-y=\dfrac{\dfrac{4b}{3}-3b}{\dfrac{10b}{3}-4b}

On further simplification, we get

h-x:k-y=\dfrac{\dfrac{4b-9b}{3}}{\dfrac{10b-12b}{3}}

h-x:k-y=\dfrac{-5b}{-2b}

h-x:k-y=\dfrac{5}{2}

h-x:k-y=5:2

Therefore, the ratio h-x:k-y is 5:2.

8 0
1 year ago
Points E, F, and D are located on circle C. Circle C is shown. Line segments E C and D F are radii. Lines are drawn from points
elena55 [62]

Answer:

Option (1). 34°

Step-by-step explanation:

From the figure attached, CE and CD are the radii of the circle C.

Central angle CED formed by the intercepted arc DE = 68°

Since measure of an arc = central angle formed by the intercepted arc

Therefore, m∠CED = 68°

Since m∠EFD = \frac{1}{2}(m\angle ECD) [Central angle of an intercepted arc measure  the double of the inscribed angle by the same arc]

Therefore, m∠EFD = \frac{1}{2}(68)

                               = 34°

Therefore, Option (1) 34° will be the answer.

7 0
2 years ago
Read 2 more answers
Polygon ABCD has sides with these lengths: , 5 units; , 4 units; , 4.5 units; and , 7 units. The slope of is 5, the slope of is
Dahasolnce [82]
Well, you can start by putting the slopes and lengths on the right side (Where is says slope of A'B'). The slopes will be the same, so Slope of AB is still 5 and Slope of BC is still 0.25. When you get to the lengths, just multiply it by 1.2. The length for Length of CD is 5.4 and Length of AD is 8.4

Here's what it should look like:
Slope of A'B' ⇔ 5
Slope of B'C' ⇔ 0.25
Length of C'D' ⇔ 5.4
Length of A'D' ⇔ 8.4
3 0
2 years ago
Read 2 more answers
Farrah borrowed $155 from her brother. She has already paid back $15. She plans to pay back $35 each month until the debt is pai
Sergeeva-Olga [200]

Answer:

4 months

Step-by-step explanation:

if she paid the 15 already subtract it from the 155. Once you that you get 140 and if you break that inton groups of 35 *division* than you get 4. As in 4 months

7 0
2 years ago
Read 2 more answers
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