All categories

3 years ago

POSSIBLE POINTS: 3.33

Mathematics

1 answer:

Triss [41]3 years ago

3 0

Answer:

From the information given, you can write the following system of equations:

x+y+z=16 (1)

5x+4.5y+5.5z=82 (2)

z=x+y (3)

You can replace (3) in (1) and (2):

x+y+x+y=16

2x+2y=16 (4)

5x+4.5y+5.5(x+y)=82

5x+4.5y+5.5x+5.5y=82

10.5x+10y=82 (5)

Now, you have the following equations:

2x+2y=16 (4)

10.5x+10y=82 (5)

You can isolate x in (4):

2x=16-2y

x=8-y (6)

Then, you have to replace (6) in (5):

10.5(8-y)+10y=82

84-10.5y+10y=82

84-82=10.5y-10y

2=0.5y

y=2/0.5

y=4

Now, you can replace the value of y in (6) to find x:

x=8-4

x=4

Finally, you can replace the vaue of x and y in (3) to find z:

z=4+4

z=8

According to this, she should buy 4 apple pies, 4 banana pies and 8 chocolate pies.

You might be interested in

In a random sample of 150 customers of a high-speed Internetprovider, 63 said that their service had been interrupted one ormore

erastovalidia [21]

Answer:

a) The 95% confidence interval would be given by (0.341;0.499)

b) The 99% confidence interval would be given by (0.316;0.524)

c) n=335

d)n=649

Step-by-step explanation:

1) Notation and definitions

$X_{IS}=63$ number of high speed internet users that had been interrupted one or more times in the past month.

$n=150$ random sample taken

$\hat p_{IS}=\frac{63}{150}=0.42$ estimated proportion of high speed internet users that had been interrupted one or more times in the past month.

$p_{IS}$ true population proportion of high speed internet users that had been interrupted one or more times in the past month.

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

1) Part a

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

$t_{\alpha/2}=-1.96, t_{1-\alpha/2}=1.96$

The confidence interval for the mean is given by the following formula:

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

If we replace the values obtained we got:

$0.42 - 1.96\sqrt{\frac{0.42(1-0.42)}{150}}=0.341$

$0.42 + 1.96\sqrt{\frac{0.42(1-0.42)}{150}}=0.499$

The 95% confidence interval would be given by (0.341;0.499)

2) Part b

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by $\alpha=1-0.99=0.01$ and $\alpha/2 =0.005$ . And the critical value would be given by:

$t_{\alpha/2}=-2.58, t_{1-\alpha/2}=2.58$

The confidence interval for the mean is given by the following formula:

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

If we replace the values obtained we got:

$0.42 - 2.58\sqrt{\frac{0.42(1-0.42)}{150}}=0.316$

$0.42 + 2.58\sqrt{\frac{0.42(1-0.42)}{150}}=0.524$

The 99% confidence interval would be given by (0.316;0.524)

3) Part c

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

And on this case we have that $ME =\pm 0.05$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

And replacing into equation (b) the values from part a we got:

$n=\frac{0.42(1-0.42)}{(\frac{0.05}{1.96})^2}=374.32$

And rounded up we have that n=335

4) Part d

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

And on this case we have that $ME =\pm 0.05$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

And replacing into equation (b) the values from part a we got:

$n=\frac{0.42(1-0.42)}{(\frac{0.05}{2.58})^2}=648.599$

And rounded up we have that n=649

5 0

3 years ago

Find the coordinates of the midpoint M and the distance between J(4,3) and K(2,-3)

Sergio [31]

Answer:

(3,0)

Step-by-step explanation:

Co-ordinates of J =(4,3)

Co-ordinates of K = (2,-3)

Midpoint formula :

= $\frac{x_{1}+x_{2} }{2} , \frac{y_{1} +y_{2} }{2} \\ \frac{4+2}{2} ,\frac{3+(-3)}{2}\\\frac{6}{2},\frac{3-3}{2}\\3, 0$

<u>Coordinates of the midpoint M </u><u>(3,0)</u>

5 0

4 years ago

Please someone please help, will mark you Brainliest

Gnoma [55]

Answer:

Step-by-step explanation:

Community gym

=> membership fee = $50

=> monthly fee = $55

=> First month = 105$

=> 50 + 55(x)

=> 50 + 55 (15)

=> 50 + 825

=> 875

Workout gym

=> membership fee = $200

=> monthly fee = $45

=> First month = 245$

=> 200 + 45(x)

=> 200 + 45(15)

=> 200 + 675

=> 875

On the 15th month, community gym and workout gym will both get the same amount of $875.

4 0

3 years ago

Read 2 more answers

Match the given angle pair to their relationship using the figure beloe

hoa [83]

Answer:

Vertical: 1

corresponding: 3

alternate interior: 2

alternate exterior: 4

6 0

3 years ago

The perimeter and the area

Genrish500 [490]

Perimeter=28/3
Area=40/3

3 0

3 years ago