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3 years ago

Whats 100,002÷13 dont try calculator or paper

Mathematics

2 answers:

balandron [24]3 years ago

6 0

100,002 divded by 13 will be 7692 when estimated

erica [24]3 years ago

5 0

7692 when estimated. Good luck

You might be interested in

What is the percent of change from 56 to 22? round to the nearest percent

V125BC [204]

$\textrm{percentage change} = \dfrac{\textrm{new} - \textrm{old}}{\textrm{old}} \times 100 \%$

$\dfrac{22 - 56}{56} \times 100\% \approx -60.7\%$

Answer: -61%

3 0

3 years ago

Work out the area of the shaded shape.

CaHeK987 [17]

The answer would be 61m^2

5x11=55
8-5=3
11-9=2
2x3=6
6+55=61

and since it is the area it is squared
hope this helps

5 0

2 years ago

Each front tire on a particular type of vehicle is supposed to be filled to a pressure of 26 psi. Suppose the actual air pressur

Ostrovityanka [42]

Answer:

1) K = 7.895 × 10⁻⁶

2) 0.3024

3) 3.6775 × 10⁻²

4) $f(x)= \frac{1}{20} +\frac{3x^{2} }{38000}$

5) X and Y are not independent variables

$h(x\mid y) = \frac{38000x^2+38000y^2}{3y^2+19000}$

7) 0.54967

8) 25.33 psi

σ = 2.875

Step-by-step explanation:

1) Here we have

$f(x, y) =\begin{cases} & \text (x^{2}+y^{2}) \right. 20\leq x\leq 30 & \ 0 \, Otherwise\end{cases}$

$\int_{x}\int_{y} f(x, y)dydx = 1$

$\int_{x}( \right )\int_{y} f(x, y)dy)dx = 1$

$K\int_{x}( \right )\int_{y}(x^{2} +y^{2})dy)dx = 1$

$K\int_{x}( (x^{2}y +\frac{y^{3}}{3})_{20}^{30})dx = 1$

$K\int_{x}( (x^{2}(30-20)) +\frac{30^{3}-20^{3}}{3})_{20}^{30})dx = 1$

$K\int_{x}( (10x^{2})+\frac{19000}{3})_{20}^{30})dx = 1$

$K( (10\frac{x^{3}}{3})+\frac{19000}{3}x)_{20}^{30})= 1$

$K( (10\frac{30^{3}-20^{3}}{3})+\frac{19000}{3}(30-20)))_{20}^{30}) = 1$

$K =\frac{3}{380000}$

2) The probability that both tires are underfilled

P(X≤26,Y≤26) =

$\int_{20}^{26} \int_{20}^{26}K(x^{2}+y^{2})dydx$

$=K\int_{x}( \right )\int_{y}(x^{2} +y^{2})dy)dx$

$= K\int_{x}( (x^{2}y +\frac{y^{3}}{3})_{20}^{26})dx$

$K\int_{x}( (x^{2}(26-20)) +\frac{26^{3}-20^{3}}{3})_{20}^{26})dx$

$K\int_{x}( (6x^{2})+\frac{9576}{3})_{20}^{26})dx$

$K( (6\frac{x^{3}}{3})+\frac{9576}{3}x)_{20}^{26})$

$K( (6\frac{26^{3}-20^{3}}{3})+\frac{9576}{3}(26-20)))_{20}^{26})$

$38304\times K =\frac{3\times38304}{380000}$

= 0.3024

That is P(X≤26,Y≤26) = 0.3024

3) The probability that the difference in air pressure between the two tires is at most 2 psi is given by

{20 ≤ x ≤ 30, 20 ≤ y ≤ 30, $| x-y |$ ≤ 2}

{20 ≤ x ≤ 30, 20 ≤ y ≤ 30, $\sqrt{(x-y)^2}$ ≤ 2}

{20 ≤ x ≤ 30, 20 ≤ y ≤ 30, y ≤ x - 2}

Which gives

20 ≤ x ≤ 22 :: 20 ≤ y ≤ x + 2

22 ≤ x ≤ 28 :: x - 2 ≤ y ≤ x + 2

28 ≤ x ≤ 30 :: x - 2 ≤ y ≤ 30

From which we derive probability as

P( $| x-y |$ ≤2) = $\int_{28}^{30} \int_{x-2}^{30}K(x^{2}+y^{2})dydx$ + $\int_{20}^{22} \int_{20}^{x+2}K(x^{2}+y^{2})dydx$ + $\int_{22}^{28} \int_{x-2}^{x+2}K(x^{2}+y^{2})dydx$

= K ( $\int_{28}^{30} \int_{x-2}^{30}K(x^{2}+y^{2})dydx$ + $\int_{20}^{22} \int_{20}^{x+2}K(x^{2}+y^{2})dydx$ + $\int_{22}^{28} \int_{x-2}^{x+2}K(x^{2}+y^{2})dydx$ )

= $K\left [ \left (\frac{14804}{15} \right )+\left (\frac{8204}{15} \right ) +\left (\frac{46864}{15} \right )\right ] = \frac{3}{380000}\times \frac{69872}{15} =\frac{4367}{118750}$ = 3.6775 × 10⁻²

4) The marginal pressure distribution in the right tire is

$f_{x}\left ( x \right )=\int_{y} f(x ,y)dy$

$=K( \right )\int_{y}(x^{2} +y^{2})dy)$

$= K( (x^{2}y +\frac{y^{3}}{3})_{20}^{30})$

$K( (x^{2}(30-20)) +\frac{30^{3}-20^{3}}{3})_{20}^{30})$

$K(10x^{2}+\frac{19000}{3})}$

$\frac{3}{38000} (10x^{2}+\frac{19000}{3})}$

$= \frac{1}{20} +\frac{3x^{2} }{38000}$

$f(x)= \frac{1}{20} +\frac{3x^{2} }{38000}$

5) Here we have

The product of marginal distribution given by

$f_x(x) f_y(y) = ( \frac{1}{20} +\frac{3x^{2} }{38000})( \frac{1}{20} +\frac{3y^{2} }{38000})$ = $\frac{(3x^2+1900)(3y^2+1900)}{1444000000}$

≠ f(x,y)

X and Y are not independent variables since the product of the marginal distribution is not joint probability distribution function.

6) Here we have the conditional probability of Y given X = x and the conditional probability of X given that Y = y is given by

$h(y\mid x) =\frac{f(x,y))}{f_{X}\left (x \right )}$ = Here we have

$h(y\mid x) =\frac{x^2+y^2}{\frac{1}{20} +\frac{3x^2}{38000} } = \frac{38000x^2+38000y^2}{3x^2+19000}$

Similarly, the the conditional probability of X given that Y = y is given by

$h(x\mid y) =\frac{x^2+y^2}{\frac{1}{20} +\frac{3y^2}{38000} } = \frac{38000x^2+38000y^2}{3y^2+19000}$

7) Here we have

When the pressure in the left tire is at least 25 psi gives

$K\int\limits^{25}_{20} \frac{38000x^2+38000y^2}{3x^2+19000} {} \, dx$

Since x = 22 psi, we have

$K\int\limits^{25}_{20} \frac{38000\cdot 25^2+38000y^2}{3\cdot 25^2+19000} {} \, dx = K \int\limits^{25}_{20} 10.066y^2+6291.39, dx = 57041.942\times \frac{3}{380000}$ = 0.45033

For P(Y≥25) we have

$K \int\limits^{30}_{25} 10.066y^2+6291.39, dx = 69624.72\times \frac{3}{380000}$ = 0.54967

8) The expected pressure is the conditional mean given by

$E(Y\mid x) = K\int\limits^{30}_{20} yh(y \mid x)\, dy$

$E(Y\mid x) = K\int\limits^{30}_{20} 10.066y^3+6291.39y\, dy = \frac{3}{380000} \times 3208609.27153$

= 25.33 psi

The standard deviation is given by

$Standard \, deviation =\sqrt{Variance}$

Variance = $K\int\limits^{30}_{20} [y-E(Y\mid x) ]^2h(y \mid x)\, dy$

$=K\int\limits^{30}_{20} [y-25.33]^2(10.066y^2+6291.39)\, dy$

= $\frac{3}{380000} \times 1047259.78$ = 8.268

The standard deviation = √8.268 = 2.875.

3 0

3 years ago

A random sample of 64 students at a university showed an average age of 20 years and a sample standard deviation of 4 years. The

adelina 88 [10]

Answer:

The 90% confidence level is $19.15< L < 20.85$

Step-by-step explanation:

From the question we are told that

The sample size is $n = 64$

The mean age is $\= x = 20 \ years$

The standard deviation is $\sigma = 4 \ years$

Generally the degree of freedom for this data set is mathematically represented as

$df = n - 1$

substituting values

$df = 64 - 1$

$df = 63$

Given that the level of confidence is 90% the significance level is mathematically evaluated as

$\alpha = 100 - 90$

$\alpha =$ 10 %

$\alpha = 0.10$

Now $\frac{\alpha }{2} = \frac{0.10}{2} = 0.05$

Since we are considering a on tail experiment

The critical value for half of this significance level at the calculated degree of freedom is obtained from the critical value table as

$t_{df, \frac{ \alpha}{2} } = t_{63, 0.05 } = 1.669$

The margin for error is mathematically represented as

$MOE = t_{df , \frac{\alpha }{2} } * \frac{\sigma}{\sqrt{n} }$

substituting values

$MOE = 1.699 * \frac{4 }{\sqrt{64} }$

$MOE = 0.85$

he 90% confidence interval for the true average age of all students in the university is evaluated as follows

$\= x - MOE < L < \= x + E$

substituting values

$20 - 0. 85 < L < 20 + 0.85$

$19.15< L < 20.85$

4 0

3 years ago

Find the first three terms of the sequence below. T n = n 2 + 3 n + 4

madam [21]

Answer:

9,14,19

Step-by-step explanation:

You seem to be giving the expression 2n+3n+4.

Now simplify

An = 5n+4

Fill in n

5(1) +4

5(2) +4

5(3)+4

After solving and simplifying you see 9, 14 and 19 are the first 3 terms.

6 0

3 years ago