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2 years ago

8

Colleen McHugh wishes to deposit checks for $13.75, $92.08, and $8.21 into her account. She would like to receive $15.00 in cash

. What is her total deposit?

2 answers:

MrRa [10]2 years ago

7 0

Her total deposit is 114.04

CaHeK987 [17]2 years ago

4 0

her total deposit is 98.94

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70 minute to 2 hours is

kondaur [170]

140 minutes ..... ok

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2 years ago

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Let x ∼ bin(9, 0.4). find

Kobotan [32]

A.
$\mathbb P(X>6)=\mathbb P(X=7)+\mathbb P(X=8)+\mathbb P(X=9)$
$=\dbinom97(0.4)^7(1-0.4)^{9-7}+\dbinom98(0.4)^8(1-0.4)^{9-8}+\dbinom99(0.4)^9(1-0.4)^{9-9}$
$\approx0.025$

b.
$\mathbb P(X\ge2)=1-\mathbb P(X$
$=1-\dbinom90(0.4)^0(1-0.4)^{9-0}-\dbinom91(0.4)^1(1-0.4)^{9-1}$
$\approx0.929$

c.
$\mathbb P(2\le X$
$=\dbinom92(0.4)^2(1-0.4)^{9-2}+\dbinom93(0.4)^3(1-0.4)^{9-3}+\dbinom94(0.4)^4(1-0.4)^{9-4}$
$\approx0.663$

d.
$\mathbb P(2$
$=\dbinom93(0.4)^3(1-0.4)^{9-3}+\dbinom94(0.4)^4(1-0.4)^{9-4}+\dbinom95(0.4)^5(1-0.4)^{9-5}$
$\approx0.669$

e.
$\mathbb P(X=0)=\dbinom90(0.4)^0(1-0.4)^{9-0}\approx0.01$

f.
$\mathbb P(X=7)=\dbinom97(0.4)^7(1-0.4)^{9-7}\approx0.021$

g, h.
For $X\sim\mathcal B(n,p)$ , recall that $\mathbb 
E[X]=\mu_X=np$ and $\mathbb V[X]={\sigma_X}^2=np(1-p)$ . So

$\mu_X=9(0.4)=3.6$
${\sigma_X}^2=9(0.4)(0.6)=2.16$

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3 years ago

Estimate the sum of 672 and 830 by rounding to the nearest hundred before adding.

Nitella [24]

Answer:

you round off to the nearest hundred which is 672 and 830 which I find to to 672 and 830 becomes 1000

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2 years ago

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An airplane flew 4 hours with a 25 mph tail wind. The return trip against the same wind took 5 hours. Find the speed of the airp

balu736 [363]

Answer:

r = 225 Mil/h speed of the airplane in still air

Step-by-step explanation:

Then:

d is traveled distance and r the speed of the airplane in still air

so the first equation is for a 4 hours trip

as d = v*t

d = 4 * ( r + 25) (1) the speed of tail wind (25 mil/h)

Second equation the trip back in 5 hours

d = 5 * ( r - 25 ) (2)

So we got a system of two equation and two unknown variables d and

r

We solve it by subtitution

from equation (1) d = 4r + 100

plugging in equation 2

4r + 100 = 5r - 125 ⇒ -r = -225 ⇒ r = 225 Mil/h

And distance is :

d = 4*r + 100 ⇒ d = 4 * ( 225) + 100

d = 900 + 100

d = 1000 miles

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3 years ago

A certain car model has a mean gas mileage of 34 miles per gallon (mpg) with a standard deviation A pizza delivery company buys

zubka84 [21]

Answer:

$z =\frac{33.3- 34}{\frac{5}{\sqrt{54}}}= -1.028$

$z =\frac{34.3- 34}{\frac{5}{\sqrt{54}}}= 0.441$

An we can use the normal standard table and the following difference and we got this result:

$P(-1.028$

Step-by-step explanation:

Assuming this statement to complete the problem "with a standard deviation 5 mpg"

We have the following info given:

$\mu = 34$ represent the mean

$\sigma= 5$ represent the deviation

We have a sample size of n = 54 and we want to find this probability:

$P(33.3 < \bar X< 34.3)$

And for this case since the sample size is large enough >30 we can apply the central limit theorem and then we can use this distribution:

$\bar X \sim N(\mu , \frac{\sigma}{\sqrt{n}})$

And we can use the z score formula given by:

$z=\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And replacing we got:

$z =\frac{33.3- 34}{\frac{5}{\sqrt{54}}}= -1.028$

$z =\frac{34.3- 34}{\frac{5}{\sqrt{54}}}= 0.441$

An we can use the normal standard table and the following difference and we got this result:

$P(-1.028$

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3 years ago