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3 years ago

Izaiah is taking a drivers

Mathematics

2 answers:

True [87]3 years ago

6 0

64 questions are on the test

yKpoI14uk [10]3 years ago

4 0

64 or 48, im pretty sure it is 64 because it doesn't say " how many are left" it says "how many questions" as in how many in total

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(a) A parachutist lands at a point on the line between the points A and B, and the target is an operation at A. The operation fa

mr Goodwill [35]

Answer:

a) $\frac{B-\frac{A+5B}{6}}{B-A}= \frac{6B -A-5B}{6(B-A)}=\frac{B-A}{6(B-A)}=\frac{1}{6}$

b) $P(X>1) = 1-P(X\leq 1)=1-\int_{0}^1 \frac{1^{2} x^{2-1} e^{- x}}{\gamma(2)}=0.736$

Step-by-step explanation:

Part a

We assume that the parachutist lands at random point in the interval (x=A,y=B) we have a continuous random variable X. And the distribution of X would be uniform $Y\sim Unif(A,B)$ . And the density function would be given by:

$f(x) =\frac{1}{B-A} , A$

And 0 for other case.

The operation fails, if parachutist's distance to A is more than five times as much as her distance to B.

So the point P in the interval (A,B) at which the distance to A is exactly 5 times the distance to B is given by:

$P= A + \frac{5}{6} (B-A)= \frac{6A +5B -5A}{6}=\frac{A+5B}{6}$

And we can find the probability desired like this:

$P(d(P,A) \geq 5 d(P,B))= P(\frac{A+5B}{6} < X< B)$

And from the cumulative distribution function of X ficen by $F(X)\frac{X-A}{B-A}$ we got:

$\frac{B-\frac{A+5B}{6}}{B-A}= \frac{6B -A-5B}{6(B-A)}=\frac{B-A}{6(B-A)}=\frac{1}{6}$

Part b

For this case we assume that $X\sim Gamma (2,1)$

On this case we assume that $\alpha=2, \beta= 1$

The density function for the Gamma distribution is given by:

$P(X)= \frac{\beta^{\alpha} x^{\alpha-1} e^{-\beta x}}{\gamma(\alpha)}$

And on this case we can find the probability using the complement rule like this:

$P(X>1) = 1-P(X\leq 1)=0.736$

We can solve this problem with the following excel code:

"=1-GAMMA.DIST(1;2;1;TRUE)"

And if we do it by hand we need to do this:

$P(X>1) = 1-P(X\leq 1)=1-\int_{0}^1 \frac{1^{2} x^{2-1} e^{- x}}{\gamma(2)}=0.736$

6 0

3 years ago

In order to estimate the average time spent on the computer terminals per student at a local university, data were collected for

Setler [38]

Answer:

7) d)

standard error of the mean of one sample of 'n' observation = 0.20

8) a)

The margin of Error = 0.392

9) d

The 95% of confidence intervals are (8.61 , 9.39)

Step-by-step explanation:

solution:-

The Given data sample size 'n' = 81

Given Population standard deviation 'σ' = 1.8 hours

The standard error of the mean of one sample of 'n' observation is

Standard error (SE)

= $\frac{S.D}{\sqrt{n} }$

= σ / √n

= $\frac{1.8}{\sqrt{81} } =0.2$

standard error of the mean of one sample of 'n' observation = 0.20

Solution:-

The Given data sample size 'n' = 81

Given Population standard deviation 'σ' = 1.8 hours

Given the probability is 0.95

The z- score = 1.96 at 0.05 level of significance.

The margin of Error = $\frac{z_{0.95} S.D}{\sqrt{n} }$

= $\frac{1.96 (S.D)}{\sqrt{n} }$

= $\frac{1.96 (1.8)}{\sqrt{81} }$

= 0.392

The margin of Error = 0.392

Solution:-

The 95% of confidence intervals are

$(x^{-} - 1.96\frac{S.D}{\sqrt{n} } , x^{-} + 1.96\frac{S.D}{\sqrt{n} } )$

$(9 - 1.96\frac{1.8}{\sqrt{81} } , 9+ 1.96\frac{1.8}{\sqrt{81} } )$

(9 - 0.392 , (9 + 0.392)

(8.609 , 9.392)

The 95% of confidence intervals are (8.61 , 9.39)

4 0

3 years ago

HELPPPPPPPFPPFPFPFFPPFPF

LiRa [457]

Answer:First doman

Step-by-step explanation:

5 0

3 years ago

Please help out explanation need it

prisoha [69]

Answer:

A=2(wl+hl+hw)

A=2(6ft×7ft+6ft×7ft+6ft×6ft)

A=2(42ft²+42ft²+36ft²)

A=2(120ft²)

A=240ft²

Step-by-step explanation:

8 0

4 years ago

Divide 6x^3 + 1 - 14x by 3x +6 solve it

Lena [83]

Answer:

$2 {x}^{2} - 4x + \frac{10}{3} - \frac{19}{3x + 6}$

3 0

4 years ago