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3 years ago

What is the elasped time for 9:30p.m to8:30a.m?

Mathematics

2 answers:

aivan3 [116]3 years ago

6 0

It would be 11 hours.

hope this helps you

solniwko [45]3 years ago

3 0

12:00-9:30 = 2:30
12:30+8:30 = 8:30
2:30 + 8:30 = 11:00
11 total hours elapsed

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Hassan just accepted a job at a new company where he will make an annual salary of

stira [4]

Step-by-step explanation:

$68 000 × 8 years

= $544 000

$1 500 × 8 years

= $12 000

$544 000 + $12 000

= $556 00

4 0

3 years ago

Determine whether the two expressions are equivalent. If so, tell what property

Aleksandr-060686 [28]

Answer:

They are equivalent by Associative Property

Step-by-step explanation:

(3•6)•9 is 18•9 which is 162

3•(6•9) is 3•54 which is also 162

So the difference between the two expressions is that they are grouped differently. In the first expression the 3•6 had to be multiplied first. In the other expression the 6•9 had to be multiplied first. But really both were 3•6•9 and it's all multiplication and can happen in any order. But the order 3, 6, 9 did not change. That is Associative property that says you can multiply by grouping factors differently and still get the same answer.

8 0

3 years ago

Randomly selected 110 student cars have ages with a mean of 8 years and a standard deviation of 3.6 years, while randomly select

monitta

Answer:

1. Yes, there is sufficient evidence to support the claim that student cars are older than faculty cars.

2. The 98% confidence interval for the difference between the two population means is [1.432 years, 3.968 years].

Step-by-step explanation:

We are given that randomly selected 110 student cars to have ages with a mean of 8 years and a standard deviation of 3.6 years, while randomly selected 75 faculty cars to have ages with a mean of 5.3 years and a standard deviation of 3.7 years.

Let $\mu_1$ = mean age of student cars.

$\mu_2$ = mean age of faculty cars.

So, Null Hypothesis, $H_0$ : $\mu_1 \leq \mu_2$ {means that the student cars are younger than or equal to faculty cars}

Alternate Hypothesis, $H_A$ : $\mu_1>\mu_2$ {means that the student cars are older than faculty cars}

(1) The test statistics that will be used here is Two-sample t-test statistics because we don't know about the population standard deviations;

T.S. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)} {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t_n_1_+_n_2_-_2$

where, $\bar X_1$ = sample mean age of student cars = 8 years

$\bar X_2$ = sample mean age of faculty cars = 5.3 years

$s_1$ = sample standard deviation of student cars = 3.6 years

$s_2$ = sample standard deviation of student cars = 3.7 years

$n_1$ = sample of student cars = 110

$n_2$ = sample of faculty cars = 75

Also, $s_p=\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(110-1)\times 3.6^{2}+(75-1)\times 3.7^{2} }{110+75-2} }$ = 3.641

So, the test statistics = $\frac{(8-5.3)-(0)} {3.641 \times \sqrt{\frac{1}{110}+\frac{1}{75} } }$ ~ $t_1_8_3$

= 4.952

The value of t-test statistics is 4.952.

Since the value of our test statistics is more than the critical value of t, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.

Therefore, we support the claim that student cars are older than faculty cars.

(2) The 98% confidence interval for the difference between the two population means ( $\mu_1-\mu_2$ ) is given by;

98% C.I. for ( $\mu_1-\mu_2$ ) = $(\bar X_1-\bar X_2) \pm (t_(_\frac{\alpha}{2}_) \times s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} })$

= $(8-5.3) \pm (2.326 \times 3.641 \times \sqrt{\frac{1}{110}+\frac{1}{75} })$

= $[2.7 \pm 1.268]$

= [1.432, 3.968]

Here, the critical value of t at a 1% level of significance is 2.326.

Hence, the 98% confidence interval for the difference between the two population means is [1.432 years, 3.968 years].

7 0

3 years ago

7(9+3) using the distributive property

ivann1987 [24]

Answer:63 plus 21

Step-by-step explanation:

6 0

3 years ago

(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