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3 years ago

What is the solution to the equation x + 7 = 63?

Mathematics

1 answer:

bezimeni [28]3 years ago

7 0

Answer:

The answer is 9.

Step-by-step explanation:

If you divide 63÷7, you'll get 9.

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The time for this event for boys in secondary school is known to possess a normal distribution with a mean of 460 seconds and a

AlekseyPX

Answer:

The time that the boys need to beat in order to earn a certificate of recognition from the fitness association is 511.264 seconds.

Step-by-step explanation:

We are given that the time for this event for boys in secondary school is known to possess a normal distribution with a mean of 460 seconds and a standard deviation of 40 seconds.

The fitness association wants to recognize the fastest 10% of the boys with certificates of recognition.

Let X = time for this event for boys in secondary school

SO, X ~ Normal( $\mu=460,\sigma^{2} =40^{2}$ )

The z-score probability distribution for normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = mean time = 460 seconds

$\sigma$ = standard deviation = 40 seconds

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, it is given that the fitness association wants to recognize the fastest 10% of the boys with certificates of recognition, which means;

P(X > x) = 0.10 {where x is the required time which boy need to beat}

P( $\frac{X-\mu}{\sigma}$ > $\frac{x-460}{40}$ ) = 0.10

P(Z > $\frac{x-460}{40}$ ) = 0.10

So, the critical value of x in the z table which represents the top 10% of the area is given as 1.2816, that is;

$\frac{x-460}{40} =1.2816$

${x-460}{} =1.2816\times 40$

$x$ = 460 + 51.264 = 511.264 seconds

Hence, the time that the boys need to beat in order to earn a certificate of recognition from the fitness association is 511.264 seconds.

8 0

3 years ago

Is 3.7 an integer Need help on my math homework plz help

Finger [1]

An integer is a number that is not a fraction so no 3.7 is not an integer

4 0

3 years ago

The lifetime X (in hundreds of hours) of a certain type of vacuum tube has a Weibull distribution with parameters α = 2 and β =

stich3 [128]

I'm assuming $\alpha$ is the shape parameter and $\beta$ is the scale parameter. Then the PDF is

$f_X(x)=\begin{cases}\dfrac29xe^{-x^2/9}&\text{for }x\ge0\\\\0&\text{otherwise}\end{cases}$

a. The expectation is

$E[X]=\displaystyle\int_{-\infty}^\infty xf_X(x)\,\mathrm dx=\frac29\int_0^\infty x^2e^{-x^2/9}\,\mathrm dx$

To compute this integral, recall the definition of the Gamma function,

$\Gamma(x)=\displaystyle\int_0^\infty t^{x-1}e^{-t}\,\mathrm dt$

For this particular integral, first integrate by parts, taking

$u=x\implies\mathrm du=\mathrm dx$

$\mathrm dv=xe^{-x^2/9}\,\mathrm dx\implies v=-\dfrac92e^{-x^2/9}$

$E[X]=\displaystyle-xe^{-x^2/9}\bigg|_0^\infty+\int_0^\infty e^{-x^2/9}\,\mathrm x$

$E[X]=\displaystyle\int_0^\infty e^{-x^2/9}\,\mathrm dx$

Substitute $x=3y^{1/2}$ , so that $\mathrm dx=\dfrac32y^{-1/2}\,\mathrm dy$ :

$E[X]=\displaystyle\frac32\int_0^\infty y^{-1/2}e^{-y}\,\mathrm dy$

$\boxed{E[X]=\dfrac32\Gamma\left(\dfrac12\right)=\dfrac{3\sqrt\pi}2\approx2.659}$

The variance is

$\mathrm{Var}[X]=E[(X-E[X])^2]=E[X^2-2XE[X]+E[X]^2]=E[X^2]-E[X]^2$

The second moment is

$E[X^2]=\displaystyle\int_{-\infty}^\infty x^2f_X(x)\,\mathrm dx=\frac29\int_0^\infty x^3e^{-x^2/9}\,\mathrm dx$

Integrate by parts, taking

$u=x^2\implies\mathrm du=2x\,\mathrm dx$

$\mathrm dv=xe^{-x^2/9}\,\mathrm dx\implies v=-\dfrac92e^{-x^2/9}$

$E[X^2]=\displaystyle-x^2e^{-x^2/9}\bigg|_0^\infty+2\int_0^\infty xe^{-x^2/9}\,\mathrm dx$

$E[X^2]=\displaystyle2\int_0^\infty xe^{-x^2/9}\,\mathrm dx$

Substitute $x=3y^{1/2}$ again to get

$E[X^2]=\displaystyle9\int_0^\infty e^{-y}\,\mathrm dy=9$

Then the variance is

$\mathrm{Var}[X]=9-E[X]^2$

$\boxed{\mathrm{Var}[X]=9-\dfrac94\pi\approx1.931}$

b. The probability that $X\le3$ is

$P(X\le 3)=\displaystyle\int_{-\infty}^3f_X(x)\,\mathrm dx=\frac29\int_0^3xe^{-x^2/9}\,\mathrm dx$

which can be handled with the same substitution used in part (a). We get

$\boxed{P(X\le 3)=\dfrac{e-1}e\approx0.632}$

c. Same procedure as in (b). We have

$P(1\le X\le3)=P(X\le3)-P(X\le1)$

and

$P(X\le1)=\displaystyle\int_{-\infty}^1f_X(x)\,\mathrm dx=\frac29\int_0^1xe^{-x^2/9}\,\mathrm dx=\frac{e^{1/9}-1}{e^{1/9}}$

Then

$\boxed{P(1\le X\le3)=\dfrac{e^{8/9}-1}e\approx0.527}$

7 0

3 years ago

An artist used silver wire to make a square that has a perimeter of 40 inches. She then used copper wire to make the largest cir

Serhud [2]

Answer:

The artist used $8.6\ in$ more of silver wire

Step-by-step explanation:

see the attached figure to better understand the problem

step 1

we know that

The Perimeter of the square is equal to

$P=4a$

where

a is the length side of the square

$P=40\ in$

$40=4a$

$a=10\ in$

Each side of the square has a length of 10 in

step 2

Find the circumference of the circle

The diameter of the circle is equal to the length of the side of the square

The circumference is equal to

$C=\pi D$

we have

$D=10\ in$

substitute

$C=(3.14)(10)=31.4\ in$

step 3

Find the difference

$40\ in-31.4\ in=8.6\ in$

4 0

4 years ago

Can you simplify 37/80

Setler [38]

Answer:

Step-by-step explanation:

37 is prime, so it's not possible to reduce/simplify 37/80. 37/80 is the ratio of two integers. We could, of course, use a calculator to find the decimal equivalent: 0.4625, but this is not more accurate than the ratio 37/80.

7 0

2 years ago