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WARRIOR [948]
2 years ago
11

T=m-n for n for the final answer

Mathematics
1 answer:
Snezhnost [94]2 years ago
6 0

Answer:n=-t+m

Step-by-step explanation:

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two cars leave Memphis from exactly the same spot. one traveling north at 70 mph and the other traveling south at 65 mph. how lo
Gemiola [76]

Answer:

2 hours

Step-by-step explanation:

270 = (70+65) * t

270= 135t

t = 2

8 0
2 years ago
Solve the triangle.<br><br> a = 10, b = 23, C = 95°
gayaneshka [121]
C² = a² + b² - (2ab * cosC) 
<span>c² = 10² + 23² - (2 * 10 * 23 * cos95) </span>
<span>c² = 100 + 529 - (460 * -.08715) </span>
<span>c² = 629 - (-40.1) </span>
<span>c² = 669.1 </span>
<span>c = 25.87 </span>

<span>(Sin C) / C = (Sin A) / A </span>
<span>(Sin 95) / 25.87 = SinA / 10, Remember 0 < A < 85 </span>
<span>(10 * Sin95) / 25.87 = Sin A </span>
<span>A = arcsin ((10 * sin95) / 25.87) </span>
<span>A = 22.65º </span>

<span>B = 180 - A - C </span>
<span>B = 180 - 95 - 22.65 </span>
<span>B = 62.35º </span>

<span>I hope this helps. Have a good day.</span>
3 0
3 years ago
One of the main contaminants of a nuclear accident, such as that at Chernobyl, is strontium-90, which decays exponentially at a
igor_vitrenko [27]

Answer: P=0.975^t


Step-by-step explanation:

We know that the general form of the exponential decay formula is


y=A(1-r)^t, where y is final amount remaining after t time, A is the original amount and r is the rate of decay


Now, the ratio of strontium-90 remaining, p , as a function of years, t , since the nuclear accident. P=\frac{A(1-0.02)^t}{A}=\frac{(0.975)^t}{1}

Hence, the ratio of remaining since the nuclear accident is P=0.975^t

3 0
3 years ago
Read 2 more answers
WILL MARK BRAINLIEST EASY 6th grade math Use the distributive property to express 36 + 16. *
Aleks04 [339]

Answer:

4 (9+4)

Step-by-step explanation:

its 4(9+4) because we are trying to make 36+16

4(8+5) is the same answer but it would be 32+20

while 4(9+4) would be 36+16

4 0
2 years ago
Read 2 more answers
According to a report from the United States Environmental Protection Agency, burning one gallon of gasoline typically emits abo
Minchanka [31]

Answer:

B. The mean amount of CO_2 emitted by the new fuel is actually lower than 89 kg but they fall to conclude it is lower than 89 kg

Step-by-step explanation:

A Type II error is the failure to reject a false null hypothesis.

Given the null and alternate hypothesis of a fuel company which wants to test a new type of gasoline designed to have lower CO_2 emissions.:

H_0: \mu = 8.9 kg\\H_a: \mu < 8.9 kg \\\text{ (where \mu is the mean amount of CO_2 emitted by burning one gallon of this new gasoline)}

where \mu is the mean amount of CO_2 emitted by burning one gallon of this new gasoline.

If the null hypothesis is false, then:

H_a: \mu < 8.9 kg

A rejection of the alternate hypothesis above will be a Type II error.

Therefore:

The Type II error is: (B) The mean amount of CO_2  emitted by the new fuel is actually lower than 89 kg but they fall to conclude it is lower than 89 kg.

8 0
3 years ago
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