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marysya [2.9K]
2 years ago
14

Need help with this assignments

Mathematics
1 answer:
Mashcka [7]2 years ago
3 0

Given values:-

P= 0.97

V= 2.85

R= 0.082

T= 295

n= ?

we have given formula,

PV=n RT

so,

n = PV/RT

n =  \frac{0.97 \times 2.85}{0.082 \times 295}  \\  =  \frac{2.7645}{24.19}  \\  = 0.1142 \\  = 0.11

0.11 is your answer.

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Which of these tables does NOT represent a function?
horrorfan [7]

Answer:

please mark me brainliest if it helps you

B does NOT represent a function.

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2 years ago
Derek uses a 136 cm flat steel bar that weighs 4 kg to make rack in the garage. Determine the weight of a 170 cm steel bar
Bumek [7]

The weight of a 170 cm steel bar will be 5 Kg

Step-by-step explanation:

Derek uses a 136 cm flat steel bar that weighs 4 kg to make rack in the garage.

1 kg = 1000 gm

So the weight of 1 cm steel bar will be  \frac{4 }{136} kg

Weight of 1 cm bar = \frac{4000}{136} gm

Let the weight of a 170 cm steel bar will be \frac{4000 }{136} × 170 gm

⇒4000 × 1.25 gm

⇒ 5000 gm

⇒5 kg

Hence, the weight of a 170 cm steel bar will be 5 Kg

4 0
3 years ago
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in 20 minutes John can rap 10 small boxes and Hank can rap 12 small boxes

Step-by-step explanation:

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2 years ago
Read 2 more answers
Ralph is leasing a $38,000 car for 36 months. The terms of his lease include an 8.5% interest rate (money factor of 0.00354) and
Alex_Xolod [135]

Answer:

hey, the answer is D 443.73

Step-by-step explanation:

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2 years ago
Trials in an experiment with a polygraph include 97 results that include 23 cases of wrong results and 74 cases of correct resul
o-na [289]

Answer:

H0:p= 0.80                 H1: p< 0.80  one tailed test

Step-by-step explanation:

We state the null and alternative hypotheses as that the results are 80 % against the claim that the results are less than 80%.

H0:p= 0.80                 H1: p< 0.80  one tailed test

p2= 0.8 , p1= 74/97= 0.763

q1= 1-0.763= 0.237    q2= 0.2

The level of significance is 0.05 .

The Z∝= ±1.645 for ∝= 0.05

The test statistic used here is

Z= p1-p2/ √pq/n

Putting the values:

Z= 0.763 -0.8 / √ 0.8*0.2/97

z= -0.037/ 0.0406

z= -0.9113

The Z∝ = ±1.645 for ∝= 0.05 for one tailed test.

As the calculated value  does not fall in the critical region  we fail to reject the null hypothesis. There is not sufficient evidence to support  the claim that such polygraph results are correct less than 80% of the time.

Using the normal probability table.

P (Z <  -0.9113)= 1- P(z= 0.9311) = 1- 0.8238= 0.1762

If P- value is smaller than the significance level reject H0.

0.1762> 0.005  Fail to reject H0.

8 0
3 years ago
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