You randomly select one card from a 52-card deck. Find the probability of selecting a red three or a black two.

Jodie made a New Year's resolution to learn how to play the violin. She bought an

stealth61 [152]

Answer:

$30

Step-by-step explanation:

If x is the cost of a lesson, the total cost of Jodie's purchase is ...

cost of lessons + rental cost = total cost

12x + 25 = 385

12x = 360 . . . . . . . subtract 25

x = 30 . . . . . . . . . . divide by 12

Each violin lesson costs $30.

6 0

2 years ago

a bag has a total of 120 notes in denominations of rs 2, rs. 5 and rs. 10. the total value of the notes in the bag is rs 760. if

Deffense [45]

So there are three variables, rs2 label x, rs5 label y, rs10 label z
Lets write equeations using these variables.
x+y+z=120 (A bag has a total of 120 notes)
2x+5y+10z=760 (total value)
2x+5*2y+10z=960 (twice as many rs5)

3 variables, 3 independent equations will give solution. There are a fair few ways to solve this.

8 0

3 years ago

The accompanying data represent the daily (for example, Monday to Tuesday) movement of Johnson & Johnson (JNJ) stock for

egoroff_w [7]

Supposing that the stock increases in 37 days, the 95% confidence interval for the proportion of days JMJ stock increases is: (0.484, 0.7292)

The lower bound is of 0.484.
The upper bound is of 0.7292.

The interpretation is that we are 95% sure that the true proportion of all days in which the JMJ stock increases is between 0.484 and 0.7292.

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $\frac{1+\alpha}{2}$ .

Supposing that it increases on 37 out of 61 days:

$n = 61, \pi = \frac{37}{61} = 0.6066$

95% confidence level

So $\alpha = 0.95$ , z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$ , so $z = 1.96$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6066 - 1.96\sqrt{\frac{0.6066(0.3934)}{61}} = 0.484$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6066 + 1.96\sqrt{\frac{0.6066(0.3934)}{61}} = 0.7292$

The 95% confidence interval for the proportion of days JMJ stock increases is (0.484, 0.7292), in which 0.484 is the lower bound and 0.7292 is the upper bound.

The interpretation is that we are 95% sure that the true proportion of all days in which the JMJ stock increases is between 0.484 and 0.7292.

A similar problem is given at brainly.com/question/16807970

4 0

2 years ago

Hii, please help me w this :(( tysm

Ipatiy [6.2K]

Answer:

7. y

=

−

1

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