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3 years ago

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A family purchased tickets for a concert and paid a total of $70. They purchased 4 child tickets and 3 adult tickets. If an adul

t ticket were twice as expensive as a child’s ticket, what was the cost of a child’s ticket? (Please write the equation)

1 answer:

Kruka [31]3 years ago

6 0

Answer:

70x+4^2+3

Step-by-step explanation:

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Find an equation of the circle that satisfies the given conditions.

hodyreva [135]

The equation of a circle:
$(x-h)^2+(y-k)^2=r^2$
(h,k) - the coordinates of the centre
r - the radius

The midpoint of the diameter is the centre of a circle.
The coordinates of the midpoint:
$(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$
(x₁,y₁), (x₂,y₂) - the coordinates of endpoints

$P(-1,1) \\
x_1=-1 \\ y_1=1 \\ \\ Q(5,9) \\ x_2=5 \\ y_2=9 \\ \\
\frac{x_1+x_2}{2}=\frac{-1+5}{2}=\frac{4}{2}=2 \\ \frac{y_1+y_2}{2}=\frac{1+9}{2}=\frac{10}{2}=5$

The centre of the circle is (2,5).

The radius is the distance between an endpoint of the diameter and the centre.
The formula for distance:
$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$(-1,1) \\ x_1=-1 \\ y_1=1 \\ \\ (2,5) \\ x_2=2 \\ y_2=5 \\ \\ d=\sqrt{(2-(-1))^2+(5-1)^2}=\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5$

The radius is 5.

$(x-2)^2+(y-5)^2=5^2 \\
\boxed{(x-2)^2+(y-5)^2=25}$

4 0

3 years ago

A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 335335 babies were born

strojnjashka [21]

Answer:

The 99% confidence interval estimate of the percentage of girls born is (74.37%, 85.63%).

Usually, 50% of the babies are girls. This confidence interval gives values considerably higher than that, so the method to increase the probability of conceiving a girl appears to be very effective.

Step-by-step explanation:

Confidence Interval for the proportion:

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 $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

$n = 335, \pi = \frac{268}{335} = 0.8$

99% confidence level

So $\alpha = 0.01$ , z is the value of Z that has a pvalue of $1 - \frac{0.01}{2} = 0.995$ , so $Z = 2.575$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.8 - 2.575\sqrt{\frac{0.8*0.2}{335}} = 0.7437$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.8 + 2.575\sqrt{\frac{0.8*0.2}{335}} = 0.8563$

For the percentage:

Multiplying the proportions by 100.

The 99% confidence interval estimate of the percentage of girls born is (74.37%, 85.63%).

Usually, 50% of the babies are girls. This confidence interval gives values considerably higher than that, so the method to increase the probability of conceiving a girl appears to be very effective.

7 0

3 years ago

Ratio equivalent to 48:24

Evgen [1.6K]

48:24 = 24:12 = 12:6 = 6:3 (divide each number by 2 or multiply by 2)

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3 years ago

Please answer this :(

Mama L [17]

A is 12.71
B is 36.93
If this is wrong I’m sorry, I really tried

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3 years ago

What is -9 -8 -4 6 in lease to greatest

bekas [8.4K]

Answe: -9, -8, -4, 6

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3 years ago