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2 years ago

Calculate δe, if q= 0.764 kj and w= -830 j . express your answer using two significant figures.

1 answer:

8 0

Given:
Q = 0.0764 kJ = 764 J, heat input
W = -830 J, useful work done

From the 1st Law of Thermodynamics, the change in internal energy is
δE = Q - W
= 764 - (-830)
= 1594 J = 1.594 kJ

Answer: 1.59 kJ (two sig. fig.)

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Answer:

The Null hypothesis represented as

$H_0: \mu_1 - \mu_2 = 0$

The Alternative hypothesis represented as

$H_a: \mu_1 - \mu_2 < 0$

p-value = P(Z < -3.37 ) = 0.000376

There is insufficient evidence to conclude that graduate students score higher, on average, on the HPI than undergraduate students

Step-by-step explanation:

From the question we are told that

The population size is $n= 650$

The sample size for graduates is $n_1 = 300$

The sample mean for graduates is $\= x _1 = 148$

The sample standard deviation for graduates is $\sigma_1 = 16$

The sample size for under-graduates is $n _2 = 350$

The sample mean for under-graduates is $\= x _2 = 153$

The sample standard deviation for graduates is $\sigma_2 = 21$

The Null hypothesis represented as

$H_0: \mu_1 - \mu_2 = 0$

The Alternative hypothesis represented as

$H_a: \mu_1 - \mu_2 < 0$

Where $\mu_1 \ and \ \mu_2$ are the population mean

Now the test statistic is mathematically represented as

$t = \frac{(\= x_1 - \= x_2 ) }{ \sqrt{ \frac{ (n_1 - 1 )\sigma_1 ^2 + (n_2 - 1)\sigma_2^2}{n_1 +n_2 -2} } * \sqrt{ \frac{1}{n_1} + \frac{1}{n_2} } }$

substituting values

$t = \frac{(148 - 153 ) }{ \sqrt{ \frac{ (300- 1 )16 ^2 + (350 - 1) 21^2}{300 +350 -2} } * \sqrt{ \frac{1}{300} + \frac{1}{350} } }$

$t = -3.37$

The p-values is mathematically evaluated as

p-value = P(Z < -3.37 ) = 0.000376

The above answer is gotten using a p-value calculator at (0.05) level of significance

Looking the p-value we see that it is less than the level of significance (0.05) so Null hypothesis is rejected

Hence there is insufficient evidence to conclude that graduate students score higher, on average, on the HPI than undergraduate students

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2 years ago

Jeanne wants to start collecting coins and orders a coin collection starter kit. The kit comes with three coins chosen at random

lesya692 [45]

Conditional probability is a measure of the probability of an event given that another event has occurred. If the event of interest is A and the event B is known or assumed to have occurred, "the conditional probability of A given B", or "the probability of A under the condition B", is usually written as P(A|B), or sometimes $P_B(A)$ .

The conditional probability of event A happening, given that event B has happened, written as P(A|B) is given by
$P(A|B)= \frac{P(A \cap B)}{P(B)}$

In the question, we were told that there are three randomly selected coins which can be a nickel, a dime or a quarter.

The probability of selecting one coin is $\frac{1}{3}$

Part A:
To find the probability that all three coins are quarters if the first two envelopes Jeanne opens each contain a quarter, let the event that all three coins are quarters be A and the event that the first two envelopes Jeanne opens each contain a quarter be B.

P(A) means that the first envelope contains a quarter AND the second envelope contains a quarter AND the third envelope contains a quarter.

Thus $P(A)= \frac{1}{3} \times \frac{1}{3} \times \frac{1}{3} = \frac{1}{27}$

P(B) means that the first envelope contains a quarter AND the second envelope contains a quarter

Thus $P(B)= \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$

Therefore, $P(A|B)=\left( \frac{ \frac{1}{27} }{ \frac{1}{9} } \right)= \frac{1}{3}$

Part B:
To find the probability that all three coins are different if the first envelope Jeanne opens contains a dime, let the event that all three coins are different be C and the event that the first envelope Jeanne opens contains a dime be D.

$P(C)= \frac{3}{3} \times \frac{2}{3} \times \frac{1}{3} = \frac{6}{27} = \frac{2}{9}$

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Therefore, $P(C|D)=\left( \frac{ \frac{2}{9} }{ \frac{1}{3} } \right)= \frac{2}{3}$

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2 years ago

For ΔABC, ∠A = 4x - 10, ∠B = 5x + 10, and ∠C = 7x + 20. If ΔABC undergoes a dilation by a scale factor of 1 3 to create ΔA'B'C'

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We know that the angles of a triangle sum to 180°. For ΔABC, this means we have:
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Combining like terms,
16x+20=180

Subtracting 20 from both sides:
16x=160

Dividing both sides by 16:
x=10
This means ∠A=4*10-10=40-10=30°; ∠B=5*10+10=50+10=60°; and ∠C=7*10+20=70+20=90.

For ΔA'B'C', we have
(2x+10)+(8x-20)+(10x-10)=180

Combining like terms,
20x-20=180

Adding 20 to both sides:
20x=200

Dividing both sides by 20:
x=10

This gives us ∠A'=2*10+10=20+10=30°; ∠B'=8*10-20=80-20=60°; and ∠C'=10*10-10=100-10=90°.

Since the angle are all congruent, ΔABC~ΔA'B'C' by AAA.

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