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erica [24]
3 years ago
15

The ratio of the number of people who own a smartphone to the number of people who own a flip phone is 4:3. If 500 more people o

wn a smartphone than a flip phone, how many people own each type of phone?
Mathematics
1 answer:
professor190 [17]3 years ago
5 0
OK. So, each 1 in your ratio is equal to 500 people. So you do 500 x 4 to get how many people have a smartphone. For people with flip phones, you need to multiply 500 x 3. So 2,000 people have smart phones and 1,500 people have flip phones.
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Suppose that a large mixing tank initially holds 100 gallons of water in which 50 pounds of salt have been dissolved. Another br
Serggg [28]

Answer:

dA/dt = 12 - 2A/(100 + t)

Step-by-step explanation:

The differential equation of this problem is;

dA/dt = R_in - R_out

Where;

R_in is the rate at which salt enters

R_out is the rate at which salt exits

R_in = (concentration of salt in inflow) × (input rate of brine)

We are given;

Concentration of salt in inflow = 4 lb/gal

Input rate of brine = 3 gal/min

Thus;

R_in = 4 × 3 = 12 lb/min

Due to the fact that solution is pumped out at a slower rate, thus it is accumulating at the rate of (3 - 2)gal/min = 1 gal/min

So, after t minutes, there will be (100 + t) gallons in the tank

Therefore;

R_out = (concentration of salt in outflow) × (output rate of brine)

R_out = [A(t)/(100 + t)]lb/gal × 2 gal/min

R_out = 2A(t)/(100 + t) lb/min

So, we substitute the values of R_in and R_out into the Differential equation to get;

dA/dt = 12 - 2A(t)/(100 + t)

Since we are to use A foe A(t), thus the Differential equation is now;

dA/dt = 12 - 2A/(100 + t)

5 0
2 years ago
A bank with a branch located in a commercial district of a city has the business objective of developing an improved process for
tatiyna

Answer:

(a) The test statistic value is -4.123.

(b) The critical values of <em>t</em> are ± 2.052.

Step-by-step explanation:

In this case we need to determine whether there is evidence of a difference in the mean waiting time between the two branches.

The hypothesis can be defined as follows:

<em>H₀</em>: There is no difference in the mean waiting time between the two branches, i.e. <em>μ</em>₁ - <em>μ</em>₂ = 0.

<em>Hₐ</em>: There is a difference in the mean waiting time between the two branches, i.e. <em>μ</em>₁ - <em>μ</em>₂ ≠ 0.

The data collected for 15 randomly selected customers, from bank 1 is:

S = {4.21, 5.55, 3.02, 5.13, 4.77, 2.34, 3.54, 3.20, 4.50, 6.10, 0.38, 5.12, 6.46, 6.19, 3.79}

Compute the sample mean and sample standard deviation for Bank 1 as follows:

\bar x_{1}=\frac{1}{n_{1}}\sum X_{1}=\frac{1}{15}[4.21+5.55+...+3.79]=4.29

s_{1}=\sqrt{\frac{1}{n_{1}-1}\sum (X_{1}-\bar x_{1})^{2}}\\=\sqrt{\frac{1}{15-1}[(4.21-4.29)^{2}+(5.55-4.29)^{2}+...+(3.79-4.29)^{2}]}\\=1.64

The data collected for 15 randomly selected customers, from bank 2 is:

S = {9.66 , 5.90 , 8.02 , 5.79 , 8.73 , 3.82 , 8.01 , 8.35 , 10.49 , 6.68 , 5.64 , 4.08 , 6.17 , 9.91 , 5.47}

Compute the sample mean and sample standard deviation for Bank 2 as follows:

\bar x_{2}=\frac{1}{n_{2}}\sum X_{2}=\frac{1}{15}[9.66+5.90+...+5.47]=7.11

s_{2}=\sqrt{\frac{1}{n_{2}-1}\sum (X_{2}-\bar x_{2})^{2}}\\=\sqrt{\frac{1}{15-1}[(9.66-7.11)^{2}+(5.90-7.11)^{2}+...+(5.47-7.11)^{2}]}\\=2.08

(a)

It is provided that the population variances are not equal. And since the value of population variances are not provided we will use a <em>t</em>-test for two means.

Compute the test statistic value as follows:

t=\frac{\bar x_{1}-\bar x_{2}}{\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}}

  =\frac{4.29-7.11}{\sqrt{\frac{1.64^{2}}{15}+\frac{2.08^{2}}{15}}}

  =-4.123

Thus, the test statistic value is -4.123.

(b)

The degrees of freedom of the test is:

m=\frac{[\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}]^{2}}{\frac{(\frac{s_{1}^{2}}{n_{1}})^{2}}{n_{1}-1}+\frac{(\frac{s_{2}^{2}}{n_{2}})^{2}}{n_{2}-1}}

   =\frac{[\frac{1.64^{2}}{15}+\frac{2.08^{2}}{15}]^{2}}{\frac{(\frac{1.64^{2}}{15})^{2}}{15-1}+\frac{(\frac{2.08^{2}}{15})^{2}}{15-1}}

   =26.55\\\approx 27

Compute the critical value for <em>α</em> = 0.05 as follows:

t_{\alpha/2, m}=t_{0.025, 27}=\pm2.052

*Use a <em>t</em>-table for the values.

Thus, the critical values of <em>t</em> are ± 2.052.

3 0
2 years ago
Round 6.512459 to the nearest ten thousandths?
PtichkaEL [24]
6.512459\\\\Since\ the\ fifth\ place\ (hundred\ thousandths)\ is\ 5\ or\ more\ (it's\ 5),\\the\ fourth\ place\ (ten\ thousandths)\ is\ increased\ by\ 1.\\\\Answer:6.512459\approx6.5125
4 0
3 years ago
Beck finished 30% of his homework in 12 minutes. How many minutes will it take him to complete all of his homework?
Bogdan [553]

Answer: The answer is 40 minutes.

Equation: If m is the total number of minutes it takes him to complete his homework, 30% of m is given by 0.3m. Since 30% was done in 12 minutes, we have 0.3m=12.

4 0
2 years ago
Please help. <br><br> Correct answer will get brainliest and full ratings.
irina1246 [14]
Hope this helps but I think the answer is A.To find this look at the points on the graph and find out what their ordered pairs are. Once you do that look at the x values for all the ordered pairs.Example the x value for the ordered pair (-3,-1) is -3.After you have found all the x values put them in order from least to greatest.I any x values are repeated then leave them out.Example if your x values are 0,1,1,2,3,3,4 then leave the extra 1 and 3 out and then you have 0,1,2,3,4.Once you have put the x values in order from least to greatest  and left out any extra repeated x values the you are done and you have your domain for the function.
5 0
2 years ago
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