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zhenek [66]
1 year ago
7

Samantha trades through a broker. On December 12, Samantha purchased stock worth $2,500. Later in the day, she sold her stock in

another
company for $2,750.
The broker charges $10 per trade. How much in brokerage fees did Samantha pay?
Samantha paid a brokerage fee of $
for the day's trades.
Mathematics
1 answer:
Eddi Din [679]1 year ago
4 0

Answer:

$20

Step-by-step explanation:

Broker's usually take a percentage of the money made or a fixed amount no matter how much is traded.

Here, the broker charges $10 per trade. So, it doesn't matter how much you do trade, the broker will charge $10 PER TRANSACTION.

Samantha bought stocks worth 2500 (1 transaction)

Samantha sold these again for 2750 (1 transaction).

So there were 2 trades made by Samantha.

So broker would charge $10 + $10 = $20

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Body armor provides critical protection for law enforcement personnel, but it does affect balance and mobility. The article "Imp
yKpoI14uk [10]

Answer:

No, there is not enough evidence to support the claim that true average task time with armor is less than 2 seconds.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that true average task time with armor is less than 2 seconds.

Then, the null and alternative hypothesis are:

H_0: \mu=2\\\\H_a:\mu< 2

The significance level is 0.01.

The sample has a size n=52.

The sample mean is M=1.95.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=0.2.

The estimated standard error of the mean is computed using the formula:

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{0.2}{\sqrt{52}}=0.028

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{1.95-2}{0.028}=\dfrac{-0.05}{0.028}=-1.803

The degrees of freedom for this sample size are:

df=n-1=52-1=51

This test is a left-tailed test, with 51 degrees of freedom and t=-1.803, so the P-value for this test is calculated as (using a t-table):

P-value=P(t

As the P-value (0.039) is bigger than the significance level (0.01), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that true average task time with armor is less than 2 seconds.

3 0
1 year ago
The speed of a sparrow is x km/h in still air. When the
Semmy [17]

Answer:

x = 5m/s

Step-by-step explanation:

Distance flying out = 12 km  (headwind)

Distance flying back = 12 km (tailwind)

total distance = 12 + 12 =24 km

wind speed = 1km/h

speed going out (with headwind) = (x - 1) km/h

speed coming back (with tailwind) = (x + 1) km/h

Time taken to go out = distance going out / speed going out

= 12 / (x-1)

Time taken to come back = distance coming back / speed coming back

= 12 / (x+1)

total time = time taken to go out + time taken to come back

5 =[ 12/(x-1) ] + [ 12/(x-1)]

expanding this, we will get

5x² - 24x - 5 = 0

solving quadratic equation, we will get

x = -1/5 (impossible because speed cannot be negative)

or

x = 5 (answer)

4 0
2 years ago
Barbara drives between Miami, Florida, and West Palm Beach, Florida. She drives 50 mi in clear weather and then encounters a thu
NeX [460]

Distance traveled in clear weather = 50 miles

Distance traveled in thunderstorm = 15 miles

Let speed in clear weather = x

⇒ Speed in thunderstorm = x-20

Total time taken for trip = 1.5 hours

We need to determine average speed in clear weather (i.e. x) and average speed in the thunderstorm (i.e. x-20 ).

Total time taken for trip = Time taken in clear weather + Time taken in thunderstorm

⇒ Total time taken for trip = \frac{Distance covered in clear weather}{Speed in clear weather} + \frac{Distance covered in thunderstorm}{Speed in thunderstorm}

⇒ 1.5 = \frac{50}{x} + \frac{15}{x-20}

⇒ 1.5 = \frac{50(x-20)+15(x)}{(x)(x-20)}

⇒ 15*x*(x-20) = 10*[50*(x-20)+15*x]

⇒ 15x² - 300x = 500x - 10,000 + 150x

⇒ 15x² - 300x = 650x - 10,000

⇒ 15x² - 950x + 10,000 = 0

⇒ 3x² - 190x + 2,000 = 0

The above equation is in the format of ax² + bx + c = 0

To determine the roots of the equation, we will first determine 'D'

D = b² - 4ac

⇒ D = (-190)² - 4*3*2,000

⇒ D = 36,100 - 24,000

⇒ D = 12,100

Now using the D to determine the two roots of the equation

Roots are: x₁ = \frac{-b+\sqrt{D}}{2a} ; x₂ = \frac{-b-\sqrt{D}}{2a}

⇒ x₁ = \frac{-(-190)+\sqrt{12,100}}{2*3} and x₂ = \frac{-(-190)-\sqrt{12,100}}{2*3}

⇒ x₁ = \frac{190+110}{6} and x₂ = \frac{190-110}{6}

⇒ x₁ = \frac{300}{6} and x₂ = \frac{80}{6}

⇒ x₁ = 50 and x₂ = 13.33

So speed in clear weather can be 50 mph or 13.33 mph. However, we know that in thunderstorm was 20 mph less than speed in clear weather.

If speed in clear weather is 13.33 mph then speed in thunderstorm would be negative, which is not possible since speed can't be negative.

Hence, the speed in clear weather would be 50 mph, and in thunderstorm would be 20 mph less, i.e. 30 mph.

7 0
2 years ago
For every $10 Glen earns, his parents
irga5000 [103]

10-2 = 8

He can spend $8

$8/$10 = 0.80

0.80 x 100 = 80 %

He can spend 80%

8 0
1 year ago
Read 2 more answers
Karson drove from his house to work at an average speed of 35 miles per hour.  The drive took him 20 minutes.  If the drive took
Nastasia [14]

Answer:

Karson's average speed on his way home was 28 miles per hour.

Step-by-step explanation:

Since Karson drove from his house to work at an average speed of 35 miles per hour, and the drive took him 20 minutes, if the drive took him 25 minutes and he used the same route in reverse, to determine what was his average speed going home, the following calculation must be performed:

60 = 35

20 = X

20 x 35/60 = X

700/60 = X

11.666 = X

25 = 11,666

60 = X

60 x 11.666 / 25 = X

27.99 = X

Therefore, Karson's average speed on his way home was 28 miles per hour.

3 0
2 years ago
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