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Brilliant_brown [7]

3 years ago

13

Read the word problem. Hershel buys 5 bags of cookies, 3 bags of granola, and several bags of fruit snacks from a bake sale. In

total, he buys 10 bags from the bake sale. How many bags of fruit snacks did Hershel buy? Which type of word problem is shown? part-whole comparison multiplication division

1 answer:

agasfer [191]3 years ago

5 0

Answer : Hershel buys 2 bags of fruit snacks.

Explanation :

Since we have given that

Number of bags of granola =3

Number of bags of cookies = 5

Total number of bags he buys from the bake sale =10

$\text{Remaining bags are of fruit snacks} =10-(5+3)=10-8=2$

So, number of bags of fruit snacks = 2

Hence, Hershel buys 2 bags of fruit snacks .

It is a comparison type of word problem.

As we just compare the items with the total items to get the remaining items.

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kobusy [5.1K]

Answer:

x = 11.25

Step-by-step explanation:

4x=45

divide 4 from both sides

4/4 x = 45/4

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marta [7]

Answer:

Brainleist is to be given to me!

Step-by-step explanation:

-2/3

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3 years ago

Let f be defined by the function f(x) = 1/(x^2+9)

riadik2000 [5.3K]

(a)

$\displaystyle\int_3^\infty \frac{\mathrm dx}{x^2+9}=\lim_{b\to\infty}\int_{x=3}^{x=b}\frac{\mathrm dx}{x^2+9}$

Substitute <em>x</em> = 3 tan(<em>t</em> ) and d<em>x</em> = 3 sec²(<em>t </em>) d<em>t</em> :

$\displaystyle\lim_{b\to\infty}\int_{t=\arctan(1)}^{t=\arctan\left(\frac b3\right)}\frac{3\sec^2(t)}{(3\tan(t))^2+9}\,\mathrm dt=\frac13\lim_{b\to\infty}\int_{t=\arctan(1)}^{t=\arctan\left(\frac b3\right)}\mathrm dt$

$=\displaystyle \frac13 \lim_{b\to\infty}\left(\arctan\left(\frac b3\right)-\arctan(1)\right)=\boxed{\dfrac\pi{12}}$

(b) The series

$\displaystyle \sum_{n=3}^\infty \frac1{n^2+9}$

converges by comparison to the convergent <em>p</em>-series,

$\displaystyle\sum_{n=3}^\infty\frac1{n^2}$

(c) The series

$\displaystyle \sum_{n=1}^\infty \frac{(-1)^n (n^2+9)}{e^n}$

converges absolutely, since

$\displaystyle \sum_{n=1}^\infty \left|\frac{(-1)^n (n^2+9)}{e^n}\right|=\sum_{n=1}^\infty \frac{n^2+9}{e^n} < \sum_{n=1}^\infty \frac{n^2}{e^n} < \sum_{n=1}^\infty \frac1{e^n}=\frac1{e-1}$

That is, ∑ (-1)ⁿ (<em>n</em> ² + 9)/<em>e</em>ⁿ converges absolutely because ∑ |(-1)ⁿ (<em>n</em> ² + 9)/<em>e</em>ⁿ| = ∑ (<em>n</em> ² + 9)/<em>e</em>ⁿ in turn converges by comparison to a geometric series.

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3 years ago

NEED ANSWER ASAP!!! Express your answer in scientific notation. 9.3 X 107 - 3.14 X 106

kirza4 [7]

Answer:

$89.86\times 10^6$

Step-by-step explanation:

$9.3\times 10^7 - 3.14\times 10^6\\$

In order to simplify the above expressions we make sure that we get same powers of 10 for both the terms.

So, we multiply and divide the first term with 10

Dividing first term by 10.

$\frac{9.3\times 10^7}{10}\\9.3\times 10^{7-1}\\9.3\times 10^6\\$

Now multiplying it by 10.

$10\times9.3\times 10^6\\93\times 10^6$

Evaluating the new expressions.

$93\times 10^6 - 3.14\times 10^6\\$

Taking $10^6$ common factor out.

$(93-3.14)\ties 10^6$

$=89.86\times 10^6$

5 0

3 years ago

Are the marks one receives in a course related to the amount of time spent studying the subject? To analyze this mysterious poss

sertanlavr [38]

Answer:

a) 98.522

b) 0.881

c) The correlation coefficient and co-variance shows that there is positive association between marks and study time. The correlation coefficient suggest that there is strong positive association between marks and study time.

Step-by-step explanation:

a.

As the mentioned in the given instruction the co-variance is first computed in excel by using only add/Sum, subtract, multiply, divide functions.

Marks y Time spent x y-ybar x-xbar (y-ybar)(x-xbar)

77 40 5.1 1.3 6.63

63 42 -8.9 3.3 -29.37

79 37 7.1 -1.7 -12.07

86 47 14.1 8.3 117.03

51 25 -20.9 -13.7 286.33

78 44 6.1 5.3 32.33

83 41 11.1 2.3 25.53

90 48 18.1 9.3 168.33

65 35 -6.9 -3.7 25.53

47 28 -24.9 -10.7 266.43

$Covariance=\frac{sum[(y-ybar)(x-xbar)]}{n-1}$

Co-variance=886.7/(10-1)

Co-variance=886.7/9

Co-variance=98.5222

The co-variance computed using excel function COVARIANCE.S(B1:B11,A1:A11) where B1:B11 contains Time x column and A1:A11 contains Marks y column. The resulted co-variance is 98.52222.

b)

The correlation coefficient is computed as

$Correlation coefficient=r=\frac{sum[(y-ybar)(x-xbar)]}{\sqrt{sum[(x-xbar)]^2sum[(y-ybar)]^2} }$

(y-ybar)^2 (x-xbar)^2

26.01 1.69

79.21 10.89

50.41 2.89

198.81 68.89

436.81 187.69

37.21 28.09

123.21 5.29

327.61 86.49

47.61 13.69

620.01 114.49

sum(y-ybar)^2=1946.9

sum(x-xbar)^2=520.1

$Correlation coefficient=r=\frac{886.7}{\sqrt{520.1(1946.9)} }$

$Correlation coefficient=r=\frac{886.7}{\sqrt{1012582.69} }$

$Correlation coefficient=r=\frac{886.7}{1006.2717 }$

$Correlation coefficient=r=0.881$

The correlation coefficient computed using excel function CORREL(A1:A11,B1:B11) where B1:B11 contains Time x column and A1:A11 contains Marks y column. The resulted correlation coefficient is 0.881.

c)

The correlation coefficient and co-variance shows that there is positive association between marks and study time. The correlation coefficient suggest that there is strong positive association between marks and study time. It means that as the study time increases the marks of student also increases and if the study time decreases the marks of student also decreases.

The excel file is attached on which all the related work is done.

7 0

3 years ago