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antiseptic1488 [7]
1 year ago
7

Damian found that 75% of the total number of teenagers surveyed like pop music. If 150 teenagers like pop music, what is the tot

al number of teenagers that Damian surveyed? Show your work
Mathematics
1 answer:
igor_vitrenko [27]1 year ago
6 0
Okay. To find the number of teens he surveyed, we can write and solve a proportion. Set it up like this: 150/x = 75/100. This is because you’re looking for the amount of students he surveyed and you know the percentage and the number of students that like pop music. 150 is over “x”, because “x” will represent the total number of students surveyed. Now, let’s cross multiply the values. 150 * 100 is 15,000. 75 * x = 75x. 15,000 = 75x. Now, divide each side by 75 to isolate the “x”. 15,000/75 is 200. x = 200. Damian surveyed a total of 200 teenagers.
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Devin borrowed $1,058 at 13 percent for nine months. What will he pay in interest?
Viktor [21]

Devin borrowed $1,058 at 13 percent for nine months.

We have to calculate the interest paid.

Interest = \frac{P \times R \times T}{100}

Substituting the values of

Principal = $1058

Rate = 13%

Time = 9 months = \frac{9}{12} year

Interest = \frac{1058 \times 13 \times 9}{12 \times 100}

Interest = 103.155

= 103.16

So, Devin will pay 103.16 as the interest.

Therefore, Option A is the correct answer.

8 0
2 years ago
Read 2 more answers
The quotient of 38 times a number and -4
nikdorinn [45]

Answer:

140y

Step-by-step explanation:

35 times a number times -4 is 35×y×(-4) which is -140y

4 0
1 year ago
Sociologists want to test whether the number of homeless people in a particular urban area is increasing. In 2010, the average n
Kobotan [32]

Answer:

The test statistic value is 15.3.

Step-by-step explanation:

The hypothesis for this test is:

<em>H</em>₀: The average number of homeless people is not increasing, i.e. <em>μ</em> = 42.3.

<em>H</em>ₐ: The average number of homeless people is increasing, i.e. <em>μ</em> > 42.3.

Given:

\bar x=45.3\\\sigma=6.2\\n=1000

As the population standard deviation is provided use a single mean <em>z</em>-test for the hypothesis testing.

The test statistic is:

z=\frac{\bar x-\mu}{\sigma/\sqrt{n}}=\frac{45.3-42.3}{6.2/\sqrt{1000}}=15.3

Thus, the test statistic value is 15.3.

7 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
n 2018, homes in East Baton Rouge (EBR) Parish sold for an average of $239,000. You take a random sample of homes in Ascension p
Olenka [21]

Answer:

Conclusion

   There is no sufficient evidence to conclude that the mean of the home prices from Ascension parish is higher than the EBR mean

Step-by-step explanation:

From the question we are told that

   The population mean for EBR is  \mu_ 1  = \$239,000

    The sample mean for Ascension parish  is \= x_2  = \$246,000

   The  p-value  is  p-value  =  0.045

     The level of significance is  \alpha = 0.01

The null hypothesis is  H_o : \mu_2  = \mu_1

The  alternative hypothesis is  H_a  :  \mu_2 > \mu_1

Here \mu_2 is the population mean for Ascension parish

   From the data given values we see that  

          p-value  >  \alpha

So we fail to reject the null hypothesis

So we conclude that there is no sufficient evidence to conclude that the mean of the home prices from Ascension parish is higher than the EBR mean

3 0
1 year ago
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