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MaRussiya [10]
4 years ago
8

A girl gave her cousin $2.15 in nicels and dimes . Altogether, she gave her cousin 28 coins . How many of them were dimes? How m

any were nickels?
Mathematics
1 answer:
mart [117]4 years ago
7 0

Answer: 15 dimes and 13 nickels

Step-by-step explanation:

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An ecological group suspects lobsters are consuming dangerous levels of mercury in a region of the world. Dangerous levels of me
Crank

Part A: 5 out 25 lobsters have dangerous levels of mercury

Part B: the 95% confidence interval is given by (4.45, 5.65)

since everything is given all I had to do is find the critical value which can be found from the resource packet = 2.064

Using the formula we find 5.05+-2.064*1.45/5 = ( 4.45,5.65)

For this case, we have 95% of confidence that the true mean for the average number of lobsters with dangerous levels of mercury is between 4.45,5.65

5 0
3 years ago
Simplify 8+3i over 4- 3i
zysi [14]

Answer:

23 + 36i over 25

Step-by-step explanation:

multiply 4+3i to the bottom and top and get 32+36i+9i^2 over 16-9i^2

then u simplify the i^2 and get 23+36i over 25

8 0
3 years ago
An engineer is going to redesign an ejection seat for an airplane. The seat was designed for pilots weighing between 150 lb and
lisov135 [29]

Answer:

(a) 0.50928

(b) 0.857685.

Step-by-step explanation:

We are given that an engineer is going to redesign an ejection seat for an airplane. The new population of pilots has normally distributed weights with a mean of 155 lb and a standard deviation of 29.2 lb i.e.;                                                         \mu = 160 lb  and \sigma = 27.5 lb

(A) We know that Z = \frac{X - \mu}{\sigma} ~ N(0,1)

Let X = randomly selected pilot  

If a pilot is randomly selected, the probability that his weight is between 150 lb and 201 lb = P(150 < X < 201)

P(150 < X < 201) = P(X < 201) - P(X <= 150)

P(X < 201) = P( \frac{X - \mu}{\sigma} < \frac{201 - 155}{29.2} ) = P(Z < 1.57) = 0.94179

P(X <= 150) = P( \frac{X - \mu}{\sigma}  < \frac{150 - 155}{29.2} ) = P(Z < -0.17) = 1 - P(Z < 0.17) = 1 - 0.56749

                                                                                                   = 0.43251

Therefore, P(150 < X < 201) = 0.94179 - 0.43251 = 0.50928 .

(B) We know that for sampling mean distribution;

           Z = \frac{Xbar - \mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

If 39 different pilots are randomly selected, the probability that their mean weight is between 150 lb and 201 lb is given by P(150 < X bar < 210);

 P(150 < X bar < 210) = P(X bar < 201) - P(X bar <= 150)

P(X bar < 201) = P( \frac{Xbar - \mu}{\frac{\sigma}{\sqrt{n} } } < \frac{201 - 155}{\frac{29.2}{\sqrt{39} } } ) = P(Z < 9.84) = 1 - P(Z >= 9.84)

                                                                                  = 0.999995

P(X bar <= 150) = P( \frac{Xbar - \mu}{\frac{\sigma}{\sqrt{n} } } < \frac{150 - 155}{\frac{29.2}{\sqrt{39} } } ) = P(Z < -1.07) = 1 - P(Z < 1.07)

                                                                                   = 1 - 0.85769 = 0.14231

Therefore,  P(150 < X bar < 210) = 0.999995 - 0.14231 = 0.857685.

C) If the tolerance level is very high to accommodate an individual pilot then it should be appropriate ton consider the large sample i.e. Part B probability is more relevant in that case.

3 0
3 years ago
Find the general solution of the following equation: y'(t) = 3y -5
Anton [14]

Answer:

The general solution of the equation is y = \frac{A}{3}e^{3t} + 5

Step-by-step explanation:

Since the differential equation is given as y'(t) = 3y -5

The differential equation is re-written as

dy/dt = 3y - 5

separating the variables, we have

dy/(3y - 5) = dt

dy/(3y - 5) = dt

integrating both sides, we have

∫dy/(3y - 5) = ∫dt

∫3dy/[3(3y - 5)] = ∫dt

(1/3)∫3dy/(3y - 5) = ∫dt

(1/3)㏑(3y - 5) = t + C

㏑(3y - 5) = 3t + 3C

taking exponents of both sides, we have

exp[㏑(3y - 5)] = exp(3t + 3C)

3y - 5 = e^{3t}e^{3C}        

3y - 5 = Ae^{3t}                A = e^{3C}

3y = Ae^{3t} + 5    

dividing through by 3, we have

y = \frac{A}{3}e^{3t} + 5

So, the general solution of the equation is y = \frac{A}{3}e^{3t} + 5

3 0
4 years ago
Match the correct y=mx+b equation to the graph: <br><br> pls show work/explanation!
11Alexandr11 [23.1K]
<h3>Answer:  y = (1/3)x - 2</h3>

Explanation:

The y intercept is -2 because the line crosses the y axis at this location. So b = -2

The slope is m = 1/3 because we go up 1 and over to the right 3 units. For instance, going from (0,-2) to (3,-1) follows this pattern.

You can use the slope formula for two points like (x1,y1) = (0,-2) and (x2,y2) = (3,-1) to find that m = 1/3

The slope formula is m = (y2-y1)/(x2-x1)

So that's how we go from y = mx+b to y = (1/3)x-2

4 0
3 years ago
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