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3 years ago

The temperature in the morning was -3.The temperature dropped 11 degrees by night. What was the temperature at night ?

Mathematics

1 answer:

Pachacha [2.7K]3 years ago

7 0

Answer:

The temperature was -14 degrees.

Step-by-step explanation:

The word "dropped" indicates that one has to subtract to the result. -3 - 11 is -14.

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Please help! first is brainliest!

Trava [24]

Answer:

compound event and independent event

Step-by-step explanation:

7 0

3 years ago

jane bought 5 3/5 boxes of raspberries and 2 1/4 boxes of strawberries. How much more boxes of raspberries did jane buy?

Veseljchak [2.6K]

Answer: 3 7/20

Step-by-step explanation:

5 3/5 - 2 1/4

Find GCD.

5 12/20 - 2 5/20

3 7/20 more boxes of raspberries than strawberries

3 0

3 years ago

A mathematics teacher wanted to see the correlation between test scores and

SpyIntel [72]

Answer:

The homework grade, to the nearest integer, for a student with a test grade of 68 is 69.

Step-by-step explanation:

The general form of the linear regression equation is:

$y=a+bx$

Here,

y = dependent variable = test grade

x = independent variable = homework grade

a = intercept

b = slope

Compute the value of a and b as follows:

$a=\frac{\sum Y\cdot \sum X^{2}-\sum X\cdot\sum XY}{n\cdot \sum X^{2}-(\sum X)^{2}}\\\\=\frac{(592\times 44909)-(591\times45227)}{(8\times44909)-(591)^{2}}\\\\=-14.316$ $b=\frac{n\cdot \sum XY-\sum X\cdot\sum Y}{n\cdot \sum X^{2}-(\sum X)^{2}}\\\\=\frac{(8\times 45227)-(591\592)}{(8\times44909)-(591)^{2}}\\\\=1.195$

The linear regression equation that represents the set of data is:

$y=-14.316+1.195x$

Compute the value of x for y = 68 as follows:

$y=-14.316+1.195x$

$68=-14.316+1.195x\\1.195x=68+14.316\\1.195x=82.316\\x=68.884\\x\approx 69$

Thus, the homework grade, to the nearest integer, for a student with a test grade of 68 is 69.

8 0

3 years ago

Find a1 and d for an arithmetic sequence with these terms. a3=-8 and a7=32

Arte-miy333 [17]

We know that
the formula for the arithmetic sequence is
an = a1 + (n - 1)*d
where
a1 is the first term
n is the numbers of terms

for n=3 a3=-8
-8=a1+(3-1)*d-----> -8=a1+2*d-----> equation 1

for n=7 a7=32
-32=a1+(7-1)*d----> -32=a1+6*d----> equation 2

multiply equation 2 by -1-----> 32=-a1-6*d------> equation 3

add equation 1 and equation 3
-8=a1+2*d
32=-a1-6*d
----------------
24=-4*d------> d=-6
-8=a1+2*(-6) (substitute the value of d in equation 1)
-8=a1-12-----> a1=-8+12-----> a1=4

the answer is
a1=4
d=-6

4 0

3 years ago

Evaluate the integral ∫2032x2+4dx. Your answer should be in the form kπ, where k is an integer. What is the value of k? (Hint: d

faltersainse [42]

Here is the correct computation of the question;

Evaluate the integral :

$\int\limits^2_0 \ \dfrac{32}{x^2 +4} \ dx$

Your answer should be in the form kπ, where k is an integer. What is the value of k?

(Hint: $\dfrac{d \ arc \ tan (x)}{dx} =\dfrac{1}{x^2 + 1}$ )

k = 4

(b) Now, lets evaluate the same integral using power series.

$f(x) = \dfrac{32}{x^2 +4}$

Then, integrate it from 0 to 2, and call it S. S should be an infinite series

What are the first few terms of S?

Answer:

(a) The value of k = 4

(b)

$a_0 = 16\\ \\ a_1 = -4 \\ \\ a_2 = \dfrac{12}{5} \\ \\a_3 = - \dfrac{12}{7} \\ \\ a_4 = \dfrac{12}{9}$

Step-by-step explanation:

(a)

$\int\limits^2_0 \dfrac{32}{x^2 + 4} \ dx$

$= 32 \int\limits^2_0 \dfrac{1}{x+4}\ dx$

$=32 (\dfrac{1}{2} \ arctan (\dfrac{x}{2}))^2__0$

$= 32 ( \dfrac{1}{2} arctan (\dfrac{2}{2})- \dfrac{1}{2} arctan (\dfrac{0}{2}))$

$= 32 ( \dfrac{1}{2}arctan (1) - \dfrac{1}{2} arctan (0))$

$= 32 ( \dfrac{1}{2}(\dfrac{\pi}{4})- \dfrac{1}{2}(0))$

$= 32 (\dfrac{\pi}{8}-0)$

$= 32 ( (\dfrac{\pi}{8}))$

$= 4 \pi$

The value of k = 4

(b) $\dfrac{32}{x^2+4}= 8 - \dfrac{3x^2}{2^1}+ \dfrac{3x^4}{2^3}- \dfrac{3x^6}{2x^5}+ \dfrac{3x^8}{2^7} -... \ \ \ \ \ (Taylor\ \ Series)$

$\int\limits^2_0 \dfrac{32}{x^2+4}= \int\limits^2_0 (8 - \dfrac{3x^2}{2^1}+ \dfrac{3x^4}{2^3}- \dfrac{3x^6}{2x^5}+ \dfrac{3x^8}{2^7} -...) dx$

$S = 8 \int\limits^2_0dx - \dfrac{3}{2^1} \int\limits^2_0 x^2 dx + \dfrac{3}{2^3}\int\limits^2_0 x^4 dx - \dfrac{3}{2^5}\int\limits^2_0 x^6 dx+ \dfrac{3}{2^7}\int\limits^2_0 x^8 dx-...$

$S = 8(x)^2_0 - \dfrac{3}{2^1*3}(x^3)^2_0 +\dfrac{3}{2^3*5}(x^5)^2_0- \dfrac{3}{2^5*7}(x^7)^2_0+ \dfrac{3}{2^7*9}(x^9)^2_0-...$

$S= 8(2-0)-\dfrac{1}{2^1}(2^3-0^3)+\dfrac{3}{2^3*5}(2^5-0^5)- \dfrac{3}{2^5*7}(2^7-0^7)+\dfrac{3}{2^7*9}(2^9-0^9)-...$

$S= 8(2-0)-\dfrac{1}{2^1}(2^3)+\dfrac{3}{2^3*5}(2^5)- \dfrac{3}{2^5*7}(2^7)+\dfrac{3}{2^7*9}(2^9)-...$

$S = 16-2^2+\dfrac{3}{5}(2^2) -\dfrac{3}{7}(2^2) + \dfrac{3}{9}(2^2) -...$

$S = 16-4 + \dfrac{12}{5}- \dfrac{12}{7}+ \dfrac{12}{9}-...$

$a_0 = 16\\ \\ a_1 = -4 \\ \\ a_2 = \dfrac{12}{5} \\ \\a_3 = - \dfrac{12}{7} \\ \\ a_4 = \dfrac{12}{9}$

6 0

3 years ago