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belka [17]
3 years ago
15

Becky adds 3/4 gallon of gasoline into her snow thrower's empty fuel tank. She uses 1/4 gallon for the snowstorm on Friday, and

then uses another 1/4 gallon for the snowstorms on Sunday. Is the fuel tank empty after becky uses the snow thrower on Sunday? Explain. If her fuel tank is not empty again, what single action could have been changed so that it is empty again? Explain.
Mathematics
1 answer:
Kaylis [27]3 years ago
8 0
Amount of fuel in the tank before the story begins . . . . . zero.

Total amount poured in . . . . . 3/4 gallon.

Total amount burned:
                   on Friday . . . . . 1/4 gallon
                   on Sunday . . . . 1/4 gallon
                        Total used . . 1/2 gallon .

Amount remaining in the tank on Monday:

               (3/4 gallon in) - (1/2 gallon burned)  =  1/4 gallon left.
                                                                        ==>  NOT empty

The tank would have been empty on Monday IF Becky
had poured in only 1/2 gallon, instead of 3/4 of a gallon
before the first flakes began to fly.
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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
6 0
3 years ago
A simple random sample of 20 days in which Parsnip ate seeds was selected, and the mean amount of time it took him to eat the se
bearhunter [10]

Answer:

(14.5-13.9) -1.69 \sqrt{\frac{3.98^2}{20}+ \frac{4.03^2}{17}} = -1.63

(14.5-13.9) +1.69 \sqrt{\frac{3.98^2}{20}+ \frac{4.03^2}{17}} = 2.83

And the confidence interval for this case -1.63 \leq \mu_1 -\mu_2 \leq 2.83

Step-by-step explanation:

We know the following info from the problem

\bar X_1 = 14.5 sample mean for the group 1

s_1 = 3.98 the standard deviation for the group 1

n_1= 20 the sample size for group 1

\bar X_2 = 13.9 sample mean for the group 2

s_1 = 4.03 the standard deviation for the group 2

n_2= 17 the sample size for group 2

We have all the conditions satisifed since we have random samples.

We want to construct a confidence interval for the true difference of means and the correct formula for this case is:

(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}

The degrees of freedom are given :

df = n_1 +n_2- 2 = 20+17-2=35

The confidence level is 0.9 or 90% and the significance level is \alpha=1-0.9=0.1 and \alpha/2 = 0.05 and the critical value for this case is:

t_{\alpha/2} = 1.69

And replacing the info given we got:

(14.5-13.9) -1.69 \sqrt{\frac{3.98^2}{20}+ \frac{4.03^2}{17}} = -1.63

(14.5-13.9) +1.69 \sqrt{\frac{3.98^2}{20}+ \frac{4.03^2}{17}} = 2.83

And the confidence interval for this case -1.63 \leq \mu_1 -\mu_2 \leq 2.83

5 0
3 years ago
The residents of a city voted on whether to raise property taxes. The ratio of yes votes to no votes was 3 to 2 . If there were
aleksandrvk [35]

Answer:

Step-by-step explanation:

4690/x = 5/2

cross multiply

(4690)*(2)=5*x

9380=5x

Divide both sides by 5.

x=1876

there were 1876 votes of "no"

(brainliest please?)

7 0
3 years ago
jingle bell jingle bell jingle bell rockJingle bell, jingle bell, jingle bell rock Jingle bells swing and jingle bells ring Snow
Kobotan [32]
Not me singing this in my head now currently dancing in my room as if it is Christmas.
7 0
3 years ago
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What is the quotient of StartFraction 7 Superscript negative 1 Baseline Over 7 Superscript negative 2 Baseline EndFraction? Star
Komok [63]

The quotient of the number given number  7 Superscript negative 1 Baseline Over 7 Superscript negative 2 Baseline is 7.

<h3>What is the quotient?</h3>

Quotient is the resultant number which is obtain by dividing a number with another. Let a number a is divided by number b. Then the quotient of these two number will be,

q=\dfrac{a}{b}

Here, (<em>a, b</em>) are the real numbers.

The number StartFraction 7 Superscript negative 1 Baseline Over 7 Superscript negative 2 Baseline EndFraction, given can be written as,

\dfrac{7^{-1}}{7^{-2}}

Let the quotient of this division is n. Therefore,

n=\dfrac{7^{-1}}{7^{-2}}

A number in numerator of a fraction with negative exponent can be written in the denominator with the same but positive exponent and vise versa. Therefore,

n=\dfrac{7^{2}}{7^{1}}\\n=7

Hence, the quotient of the number given number  7 Superscript negative 1 Baseline Over 7 Superscript negative 2 Baseline is 7.

Learn more about the quotient here;

brainly.com/question/673545

7 0
2 years ago
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