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liberstina [14]

3 years ago

7

Brian decided to start a dog-walking service. He's going to charge each dog owner $4.50 to walk one dog and $6.75 to walk two do

gs. Approximately how much will he earn if he walks 13 single dogs and 9 sets of two dogs?

1 answer:

puteri [66]3 years ago

8 0

Answer:

119.25

Step-by-step explanation:

13 x 4.50 = 58.5

9 x 6.75 = 60.75

58.5 + 60.75 = 119.25

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Kirti knows the following information from a study on cold medicine that included 60 participants: 30 participants in total rece

GalinKa [24]

The frequency table is attached.

Since there were 30 people out of the 60 that received cold medicine, that means that 60-30=30 people did not.

Since 26 people had a cold longer than 7 days, this means 60-26=34 had a cold that was less than or equal to 7 days.

14 people had a cold longer than 7 days and received medicine; this means that 26-14=12 people had a cold longer than 7 days and did not receive medicine.

30-14=16 people received medicine and had a cold less than or equal to 7 days.

30-12=18 people did not receive medicine and had a cold less than or equal to 7 days.

4 0

4 years ago

Describe the pattern 2, 4, 9, 11, 16, 18, 23

s2008m [1.1K]

The pattern is 2+ and 5+

7 0

3 years ago

Read 2 more answers

In a sample of 42 water specimens taken from a construction site, 26 contained detectable levels of lead. Construct a 95% conden

S_A_V [24]

Answer:

The 95% confidence interval for the proportion of water specimens that contain detectable levels of lead is (0.472,0.766).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

In a sample of 42 water specimens taken from a construction site, 26 contained detectable levels of lead.

This means that $n = 42, \pi = \frac{26}{42} = 0.619$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.619 - 1.96\sqrt{\frac{0.619*0.381}{42}} = 0.472$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.619 + 1.96\sqrt{\frac{0.619*0.381}{42}} = 0.766$

The 95% confidence interval for the proportion of water specimens that contain detectable levels of lead is (0.472,0.766).

4 0

3 years ago

Solve the equation. Y/8+4=13

Troyanec [42]

Answer:

Y =72

Step-by-step explanation:

Y/8+4=13

Subtract 4 from each side

Y/8+4-4=13-4

Y/8 = 9

Multiply by 8

Y/8*8 = 9*8

Y =72

7 0

3 years ago

Read 2 more answers

One mole of ammonia gas is contained in a vessel which is capable of changing its volume (a compartment sealed by a piston, for

artcher [175]

Answer:

a) Option C is correct.

It decreases slightly.

b) Option A is correct.

It increases slightly.

c) dU = 108.986 J = 109 J

Step-by-step explanation:

dU = 840dV + 27.32dT

With U = Total Energy of Ammonia in Joules

V = volume of ammonia (in cubic meters)

T = Temperature of Ammonia in Kelvin.

a) How does the energy change if the volume is held constant and the temperature is decreased slightly?

dU = 840dV + 27.32dT

If the volume is held constant, dV = 0

Then the temperature is decreased slightly, dT = -x (where x is a small number; minus sign indicates a decrease in temperature)

dU = 840(0) + 27.32 (-x)

dU = -27.32x

Obviously a slight decrease in total energy; due to the minus sign. A negative change indicates decrease.

b) How does the energy change if the temperature is held constant and the volume is increased slightly?

dU = 840dV + 27.32dT

If the temperature is held constant, dT = 0

Then the volume is increased slightly, dV = x (where x is a very small number)

dU = 840(x) + 27.32 (0)

dU = 840x

Obviously a slight increase in total energy; due to the positive sign. A positive change indicates increase.

c) Find the approximate change in energy if the gas is compressed by 350 cubic centimeters and heated by 4 degrees Kelvin.

The ammonia gas is compressed by 350 cm³

In the units for the equation,

350 cm³ = (350 × 10⁻⁶) m³ = (3.50 × 10⁻⁴) m³

So, the ammonia gas is compressed by (3.50 × 10⁻⁴) m³. A compression is a decrease in volume, hence,

dV = - (3.50 × 10⁻⁴) m³

The ammonia gas is heated by 4 K.

Heating indicates an increase in temperature, hence,

dT = 4 K

dU = 840dV + 27.32dT

dU = 840 (-3.50 × 10⁻⁴) + 27.32 (4)

dU = -0.294 + 109.28 = 108.986 J ≈ 109 J

Hope this Helps!!!

3 0

3 years ago