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1 year ago

Solve: 3.2/s = 0.4/8 640 64 6.4 0.16

Mathematics

2 answers:

Harrizon [31]1 year ago

3 0

Answer:6

Step-by-step explanation:

balandron [24]1 year ago

3 0

Step-by-step explanation:

$\frac{3.2}{s} = \frac{0.4}{8}$

cross multiply ✖

$3.2 \times 8 = 0.4s$

$0.4s = 25.6$

$s = 64$

$s = \frac{25.6}{0.4}$

Hope it will help you.

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A neighborhood is trying to set up school carpools, but they need to determine the number of students that need to travel to the

katrin [286]

Answer:

The correct option is 1.

Step-by-step explanation:

From the given histogram it is clear that

In age group 5-10, the number of students is 7.

In age group 11-13, the number of students is 5.

In age group 14-18, the number of students is 4.

It means age of 7 students lie in the interval 5-10, age of 5 students lie in the interval 11-13 and age of 4 students lie in the interval 14-18.

Total number of elements in the data set is $7+5+4=16$

In option 3 and 4 number of elements is data set is less than 16. So, option 3 and 4 are incorrect.

In option 2 all the data set lie in the interval 5-10. So, option 2 is incorrect.

In option 1,

Class intervals elements frequency

5-10 5,6,7,8,9,10,10 7

11-13 11,11,12,12,13 5

14-18 14,14,15,16 4

From the above table we can conclude that data set 1 represented in the given histogram.

Therefore the correct option is 1.

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1 year ago

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There are 160 deer on Manitow Island. The population of deer is growing by approximately 20% per year. Write an equation using d

soldier1979 [14.2K]

I believe the equation would be
160+0.2y

Because 160 deer and you round the 20% into a decimal point.

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1 year ago

Loretta deposits $350 every quarter into a savings account that earns 4.5% interest compounded quarterly. What is the balance af

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I am thinking the answer is C

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Use Simpson's Rule with n = 10 to estimate the arc length of the curve. Compare your answer with the value of the integral produ

SOVA2 [1]

$y=\ln(6+x^3)\implies y'=\dfrac{3x^2}{6+x^3}$

The arc length of the curve is

$\displaystyle\int_0^5\sqrt{1+\frac{9x^4}{(6+x^3)^2}}\,\mathrm dx$

which has a value of about 5.99086.

Let $f(x)=\sqrt{1+\frac{9x^4}{(6+x^3)^2}}$ . Split up the interval of integration into 10 subintervals,

[0, 1/2], [1/2, 1], [1, 3/2], ..., [9/2, 5]

The left and right endpoints are given respectively by the sequences,

$\ell_i=\dfrac{i-1}2$

$r_i=\dfrac i2$

with $1\le i\le10$ .

These subintervals have midpoints given by

$m_i=\dfrac{\ell_i+r_i}2=\dfrac{2i-1}4$

Over each subinterval, we approximate $f(x)$ with the quadratic polynomial

$p_i(x)=f(\ell_i)\dfrac{(x-m_i)(x-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+f(m_i)\dfrac{(x-\ell_i)(x-r_i)}{(m_i-\ell_i)(m_i-r_i)}+f(r_i)\dfrac{(x-\ell_i)(x-m_i)}{(r_i-\ell_i)(r_i-m_i)}$

so that the integral we want to find can be estimated as

$\displaystyle\sum_{i=1}^{10}\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx$

It turns out that

$\displaystyle\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx=\frac{f(\ell_i)+4f(m_i)+f(r_i)}6$

so that the arc length is approximately

$\displaystyle\sum_{i=1}^{10}\frac{f(\ell_i)+4f(m_i)+f(r_i)}6\approx5.99086$

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