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2 years ago

2.) Bob has two weekend jobs. Last weekend he made a total of $77 after

Mathematics

2 answers:

RUDIKE [14]2 years ago

5 0

Answer:

cashier:9 and deliverring newspapers :8

Step-by-step explanation:

aalyn [17]2 years ago

3 0

Answer:

Step-by-step explanation:

x= money earned per hour working as a cashier.

y=money earned per hour working delivering newspapers.

We propose the following system of equations:

5x + 4y =77

6x + 3y=78

We solve the system by reduction method:

-6*(5x+4y=77) ⇒ -30x-24y=-462

5*(6x+3y=78)⇒ 30x+15y=390

---------------------------

-9y=-72 ⇒ y=-72 /-9=8

Now, we get the value of "x" replacing the value of "y" by "8" in any part of the equation above.

5x + 4(8)=77

5x+32=77

5x=77-32

5x=45

x=45/5=9

therefore;

money earned per hour working as a cashier= $9/ hour

money earned per hour working delivering newspapers=$8/hour

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According to a study in a medical journal, 202 of a sample of 5,990 middle-aged men had developed diabetes. It also found that m

tekilochka [14]

Answer:

0.0588 = 5.88% probability that a middle-aged man with diabetes is very active

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Has diabetes.

Event B: Is very active.

Probability of having diabetes:

To find this probability, we take in consideration that:

It also found that men who were very active (burning about 3,500 calories daily) were a fourth as likely to develop diabetes compared with men who were sedentary. Assume that one-fifth of all middle-aged men are very active, and the rest are classified as sedentary.

So the probability of developing diabetes is:

x of 4/5 = x of 0.8(not active)

x/4 = 0.25x of 1/5 = 0.2(very active). So

$P(A) = 0.8x + 0.25*0.2x = 0.85x$

Probability of developing diabetes while being very active:

0.25x of 0.2. So

$P(A \cap B) = 0.25x*0.2 = 0.05x$

What is the probability that a middle-aged man with diabetes is very active?

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.05x}{0.85x} = \frac{0.05}{0.85} = 0.0588$

0.0588 = 5.88% probability that a middle-aged man with diabetes is very active

4 0

2 years ago

Joaquin and Trisha are playing a game in which the lower median wins the game. Their scores are shown below.

iragen [17]

Median is defined as the middle term when the scores are arranged in increasing/decreasing order.

Joaquin's scores: 58, 62, 72, 75, 85, 91 Median: 73.5
Trisha's scores: 55, 74, 76, 90, 91, 92 Median: 83

The answer is letter A.

3 0

1 year ago

Read 2 more answers

Find the mass and center of mass of the lamina that occupies the region D and has the given density function rho. D = {(x, y) |

Bas_tet [7]

Answer:

$M=168k$

$(\bar{x},\bar{y})=(5,\frac{85}{28})$

Step-by-step explanation:

Let's begin with the mass definition in terms of density.

$M=\int\int \rho dA$

Now, we know the limits of the integrals of x and y, and also know that ρ = ky², so we will have:

$M=\int^{9}_{1}\int^{4}_{1}ky^{2} dydx$

Let's solve this integral:

$M=k\int^{9}_{1}\frac{y^{3}}{3}|^{4}_{1}dx$

$M=k\int^{9}_{1}21dx$

$M=21k\int^{9}_{1}dx=21k*x|^{9}_{1}$

So the mass will be:

$M=21k*8=168k$

Now we need to find the x-coordinate of the center of mass.

$\bar{x}=\frac{1}{M}\int\int x*\rho dydx$

$\bar{x}=\frac{1}{M}\int^{9}_{1}\int^{4}_{1}x*ky^{2} dydx$

$\bar{x}=\frac{k}{168k}\int^{9}_{1}\int^{4}_{1}x*y^{2} dydx$

$\bar{x}=\frac{1}{168}\int^{9}_{1}x*\frac{y^{3}}{3}|^{4}_{1}dx$

$\bar{x}=\frac{1}{168}\int^{9}_{1}x*21 dx$

$\bar{x}=\frac{21}{168}\frac{x^{2}}{2}|^{9}_{1}$

$\bar{x}=\frac{21}{168}*40=5$

Now we need to find the y-coordinate of the center of mass.

$\bar{y}=\frac{1}{M}\int\int y*\rho dydx$

$\bar{y}=\frac{1}{M}\int^{9}_{1}\int^{4}_{1}y*ky^{2} dydx$

$\bar{y}=\frac{k}{168k}\int^{9}_{1}\int^{4}_{1}y^{3} dydx$

$\bar{y}=\frac{1}{168}\int^{9}_{1}\frac{y^{4}}{4}|^{4}_{1}dx$

$\bar{y}=\frac{1}{168}\int^{9}_{1}\frac{255}{4}dx$

$\bar{y}=\frac{255}{672}\int^{9}_{1}dx$

$\bar{y}=\frac{255}{672}8=\frac{2040}{672}$

$\bar{y}=\frac{85}{28}$

Therefore the center of mass is:

$(\bar{x},\bar{y})=(5,\frac{85}{28})$

I hope it helps you!

3 0

2 years ago

Given that the two parallel lines are cut by a transversal, and the measure of angle 2 is 50 degrees. Find each of the missing a

Mandarinka [93]

Answer:

Step-by-step explanation:

Lines m and l are the parallel lines and a line 'n' is a transverse intersecting these lines.

m∠2 = 50°

m∠1 + m∠2 = 180° [Linear pair of angles]

m∠1 = 180° - 50°

m∠1 = 130°

m∠3 = m∠1 = 130° [Vertically opposite angles]

m∠3 + m∠5 = 180° [Consecutive interior angles]

m∠5 = 180° - m∠3

= 180° - 130°

= 50°

m∠6 + m∠5 = 180° [Linear pair of angles]

m∠6 = 180° - 50° = 130°

7 0

2 years ago

Need help asap. Consider the diagram.

fgiga [73]

ANSWER

<em>alternate interior angles theorem</em>

EXPLANATION

According to the alternate interior angles theorem, when two parallel lines are are intercepted by a straight line (transversal) the angles in the interior corners of a Z-shape pattern are congruent.

From the above diagram line r is parallel to line s, therefore

$\angle \: 3 \cong \angle6$

and

$\angle \: 4\cong \angle5$

because they are alternate interior angles.

See attachment for how to spot alternate interior angles.

5 0

1 year ago

Read 2 more answers