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2 years ago

Whats a mode in math

Mathematics

2 answers:

Elza [17]2 years ago

8 0

Mode: Number that occurs most often in a set of data.

For example, 6.5 and 6.2 is the mode for this set of data because these two numbers appear more than any other number.

Eva8 [605]2 years ago

7 0

The number that appears in most often set of numbers, hope this helps ❤

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Among 46- to 51-year-olds, 28% say they have called a talk show while under the influence of peer pressurepeer pressure. Supp

sasho [114]

Answer:

0.8997, 0.9999

Step-by-step explanation:

Given that among 46- to 51-year-olds, 28% say they have called a talk show while under the influence of peer pressure.

i.e. X no of people who say they have called a talk show while under the influence of peer pressure is binomial with p = 0.28

Each person is independent of the other and there are only two outcomes

n =7

a) the probability that at least one has not called a talk showcalled a talk show while under the influence of peer pressure

= $P(Y\geq 1)$ where Y is binomial with p = 0.72

= $1-(1-0.72)^7\\=0.9999$

b) the probability that at least one has called a talk showcalled a talk show while under the influence of peer pressurepeer pressurethe probability that at least one has not called a talk showcalled a talk show while under the influence of peer pressure

= $P(x\geq 1) =1-P(0)\\= 1-(1-0.28)^7\\=0.8997$

3 0

2 years ago

70 points

Verizon [17]

Answer:

500

All of the above

paying your balances off in full

make credit card purchases that you know you can pay back within a month

0% interest for 1 year and 12% interest after that

paying your balances off in full

all of the above

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

How are unit rates, constant of proportionality, and scale factors are related.

vfiekz [6]

I found this online I hope it helps a little if not, then sorry.

6 0

2 years ago

If X and Y are independent continuous positive random

Leni [432]

a) $Z=\frac XY$ has CDF

$F_Z(z)=P(Z\le z)=P(X\le Yz)=\displaystyle\int_{\mathrm{supp}(Y)}P(X\le yz\mid Y=y)P(Y=y)\,\mathrm dy$

$F_Z(z)\displaystyle=\int_{\mathrm{supp}(Y)}P(X\le yz)P(Y=y)\,\mathrm dy$

where the last equality follows from independence of $X,Y$ . In terms of the distribution and density functions of $X,Y$ , this is

$F_Z(z)=\displaystyle\int_{\mathrm{supp}(Y)}F_X(yz)f_Y(y)\,\mathrm dy$

Then the density is obtained by differentiating with respect to $z$ ,

$f_Z(z)=\displaystyle\frac{\mathrm d}{\mathrm dz}\int_{\mathrm{supp}(Y)}F_X(yz)f_Y(y)\,\mathrm dy=\int_{\mathrm{supp}(Y)}yf_X(yz)f_Y(y)\,\mathrm dy$

b) $Z=XY$ can be computed in the same way; it has CDF

$F_Z(z)=P\left(X\le\dfrac zY\right)=\displaystyle\int_{\mathrm{supp}(Y)}P\left(X\le\frac zy\right)P(Y=y)\,\mathrm dy$

$F_Z(z)\displaystyle=\int_{\mathrm{supp}(Y)}F_X\left(\frac zy\right)f_Y(y)\,\mathrm dy$

Differentiating gives the associated PDF,

$f_Z(z)=\displaystyle\int_{\mathrm{supp}(Y)}\frac1yf_X\left(\frac zy\right)f_Y(y)\,\mathrm dy$

Assuming $X\sim\mathrm{Exp}(\lambda_x)$ and $Y\sim\mathrm{Exp}(\lambda_y)$ , we have

$f_{Z=\frac XY}(z)=\displaystyle\int_0^\infty y(\lambda_xe^{-\lambda_xyz})(\lambda_ye^{\lambda_yz})\,\mathrm dy$

$\implies f_{Z=\frac XY}(z)=\begin{cases}\frac{\lambda_x\lambda_y}{(\lambda_xz+\lambda_y)^2}&\text{for }z\ge0\\0&\text{otherwise}\end{cases}$

and

$f_{Z=XY}(z)=\displaystyle\int_0^\infty\frac1y(\lambda_xe^{-\lambda_xyz})(\lambda_ye^{\lambda_yz})\,\mathrm dy$

$\implies f_{Z=XY}(z)=\lambda_x\lambda_y\displaystyle\int_0^\infty\frac{e^{-\lambda_x\frac zy-\lambda_yy}}y\,\mathrm dy$

I wouldn't worry about evaluating this integral any further unless you know about the Bessel functions.

6 0

3 years ago

A jet flies at a rate of 1.3 x 10^ 6 feet per hour. Written in scientific notation, which is the best estimate of how many feet

nadezda [96]

(1.3X10^6ft/hr)(2.8X10^3hr)

(1.3*2.8)(10^6*10^3)

Rule (a^b)(a^c)=a^(b+c)

(3.64)(10^(6+3))

3.64(10^9)

3.64X10^9 ft

Technically we only had two significant figures and the answer should be:

3.6X10^9 if we were to express our answer to the correct number of significant figures....

7 0

3 years ago

Read 2 more answers