What is the length of bc?

Use the system of equations to answer the question. 2x+3y=8 x-y=5

Digiron [165]

Answer:

x=5+y

2(5+y)+3y=8

10+2y+3y=8

2y+3y=8-10

5y= -2

y= -2/5

x-(-2/5)=5

x+2/5=5

x=5-2/5

x=23/5

7 0

3 years ago

At the local clothing store 3 similar shirts and 4 similar jackets cost $360. 1 shirt and 3 jackets cost $220. Find the cost of

notsponge [240]

To solve this problem, it is easiest to set up a system of equations. Let's let the variable s represent the cost of a shirt and the variable j represent the cost of jackets. According to the given information, we can set up the following equations (because cost multiplied by quantity yields price):

3s + 4j = 360

1s + 3j = 220

Next, we can manipulate the second equation so that it equals s in terms of j. We do this by subtracting 3j from both sides of the equation, as shown below:

s = 220 - 3j

After that, we should substitute in this value for the variable s in the first equation.

3(220-3j) + 4j = 360

Next, we should use the distributive property to simplify the left side of the equation.

660 - 9j + 4j = 360

Then, we should simplify the left side of the equation by combining like terms.

660 - 5j = 360

After, we can subtract 660 from both sides of the equation to get the variable term alone.

-5j = -300

Finally, we should divide both sides of the equation by -5 in order to get the variable j alone.

j = 60

Now that we know the value of the variable j, we should substitute this value into one of the original equations and solve using division and subtraction to isolate the variable.

3s + 4j = 360

3s + 4(60) = 360

3s + 240 = 360

3s = 120

s = 40

Therefore, the cost of one shirt is $40.

Hope this helps!

7 0

3 years ago

According to an NRF survey conducted by BIGresearch, the average family spends about $237 on electronics (computers, cell phones

Usimov [2.4K]

Answer:

(a) Probability that a family of a returning college student spend less than $150 on back-to-college electronics is 0.0537.

(b) Probability that a family of a returning college student spend more than $390 on back-to-college electronics is 0.0023.

(c) Probability that a family of a returning college student spend between $120 and $175 on back-to-college electronics is 0.1101.

Step-by-step explanation:

We are given that according to an NRF survey conducted by BIG research, the average family spends about $237 on electronics in back-to-college spending per student.

Suppose back-to-college family spending on electronics is normally distributed with a standard deviation of $54.

Let X = back-to-college family spending on electronics

SO, X ~ Normal( $\mu=237,\sigma^{2} =54^{2}$ )

The z score probability distribution for normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean family spending = $237

$\sigma$ = standard deviation = $54

(a) Probability that a family of a returning college student spend less than $150 on back-to-college electronics is = P(X < $150)

P(X < $150) = P( $\frac{X-\mu}{\sigma}$ < $\frac{150-237}{54}$ ) = P(Z < -1.61) = 1 - P(Z $\leq$ 1.61)

= 1 - 0.9463 = 0.0537

The above probability is calculated by looking at the value of x = 1.61 in the z table which has an area of 0.9463.

(b) Probability that a family of a returning college student spend more than $390 on back-to-college electronics is = P(X > $390)

P(X > $390) = P( $\frac{X-\mu}{\sigma}$ > $\frac{390-237}{54}$ ) = P(Z > 2.83) = 1 - P(Z $\leq$ 2.83)

= 1 - 0.9977 = 0.0023

The above probability is calculated by looking at the value of x = 2.83 in the z table which has an area of 0.9977.

(c) Probability that a family of a returning college student spend between $120 and $175 on back-to-college electronics is given by = P($120 < X < $175)

P($120 < X < $175) = P(X < $175) - P(X $\leq$ $120)

P(X < $175) = P( $\frac{X-\mu}{\sigma}$ < $\frac{175-237}{54}$ ) = P(Z < -1.15) = 1 - P(Z $\leq$ 1.15)

= 1 - 0.8749 = 0.1251

P(X < $120) = P( $\frac{X-\mu}{\sigma}$ < $\frac{120-237}{54}$ ) = P(Z < -2.17) = 1 - P(Z $\leq$ 2.17)

= 1 - 0.9850 = 0.015

The above probability is calculated by looking at the value of x = 1.15 and x = 2.17 in the z table which has an area of 0.8749 and 0.9850 respectively.

Therefore, P($120 < X < $175) = 0.1251 - 0.015 = 0.1101

5 0

4 years ago

What are the diameter and circumference of the unit circle? Answer using π.

sveticcg [70]

Answer:

Diameter = 2 Circumference = 2π

Step-by-step explanation:

The diameter of a circle is the radius*2 and the radius of a unit circle is 1 so the diameter is 2. The circumference of a circle is the diameter*π. Since you are allowed to express it in terms of π. Just leave it at 2π, though the actual value of π is approximately 3.14

4 0

4 years ago

Find the volume of the following solid. Using cylindrical coordinates

Reika [66]

In cylindrical coordinates, the equations of the surfaces become

$z = 4 - 4r^2 \\\\ z = (r^2)^2 - 1 = r^4 - 1$

These surfaces intersect on the cylinder of radius 1 with cross sections parallel to the x,y-plane:

$4 - 4r^2 = r^4 - 1 \\\\ \implies r^4 + 4r^2 - 5 = (r-1)(r+1)(r^2+5) = 0 \\\\ \implies r=1$

Then in cylindrical coordinates, the volume of the space bounded by these surfaces is

$\displaystyle \int_0^{2\pi}\int_0^1\int_{r^4-1}^{4-4r^2}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta = 2\pi \int_0^1 \int_{r^4 - 1}^{4-4r^2}r\,\mathrm dz\,\mathrm dr \\\\ = \pi \int_0^1 (4-4r^2)^2 - (r^4-1)^2 \,\mathrm dr \\\\ = \pi \int_0^1 (15-32r^2+18r^4-r^8)\,\mathrm dr = \boxed{\frac{352\pi}{45}}$

7 0

3 years ago