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2 years ago

9

Select the correct answer. what is the probability of randomly picking an ace of hearts from a standard deck of playing cards? a

ssume there are no jokers in the deck. a. 1/52 b. 3/52 c. 1/13 d. 1/4

2 answers:

TiliK225 [7]2 years ago

8 0

A. Hope this helps!!!

Vladimir79 [104]2 years ago

5 0

A.1/52 since their is only one ace of hearts and no jokers

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It took emily 25 min to rideHer bicycle to the repair shop in one hour and 15 minutes to walk back home. If Emily her bicycle to

s2008m [1.1K]

Answer:the repair shop is 5km from her house.

Step-by-step explanation:

Let x represent the distance of the repair shop from her house.

Let y represent Emily's speed when riding.

Distance = speed × time

It took Emily 25 min to ride her bike to the repair shop. Converting 25 min to hours, it becomes 25/60 = 5/12 hours.

Distance covered,

x = y × 5/12 = 5y/12

It took her 1h 15 min to walk back home. Converting to hours, it becomes 1 + 15/60 = 1 1/4 = 5/4

If she can ride her bike 8km/h faster than she can walk, it means that her speed while walking would be y 8 8

Therefore,

Distance covered,

x= 5/4(y - 8) = (5y - 40)/4

Since the distance remains the same, then

5y/12 =(5y - 40)/4

Crossmultiplying, it becomes

5y × 4 = 12(5y - 40)

20y = 60y - 480

60y - 20y = 480

40y = 480

y = 480/40 = 12

x = 5y/12 = 5 × 12/12 = 5 km

6 0

3 years ago

Prove that H c G is a normal subgroup if and only if every left coset is a right coset, i.e., aH = Ha for all a e G

Kaylis [27]

$\Rightarrow$

Suppose first that $H\subset G$ is a normal subgroup. Then by definition we must have for all $a\in H$ , $xax^{-1} \in H$ for every $x\in G$ . Let $a\in G$ and choose $(ab)\in aH$ ( $b\in H$ ). By hypothesis we have $aba^{-1} =abbb^{-1}a^{-1}=(ab)b(ab)^{-1} \in H$ , i.e. $aba^{-1}=c$ for some $c\in H$ , thus $ab=ca \in Ha$ . So we have $aH\subset Ha$ . You can prove $Ha\subset aH$ in the same way.

$\Leftarrow$

Suppose $aH=Ha$ for all $a\in G$ . Let $h\in H$ , we have to prove $aha^{-1} \in H$ for every $a\in G$ . So, let $a\in G$ . We have that $ha^{-1} =a^{-1}h'$ for some $h'\in H$ (by the hypothesis). hence we have $aha^{-1}=h' \in H$ . Because $a$ was chosen arbitrarily we have the desired .

5 0

3 years ago

A rental car company charges $31.91 per day to rent a car and $0.07 for every mile driven. Hudson wants to rent a car, knowing t

qaws [65]

The inequality that can be used to determine x is $31.91x + 7 \leq 70$

<u />

<h3>How to generate the inequality</h3>

<u>Given data</u>

Rental company charge per day = $31.91

Rental company charge per mile driven = $0.07

Hudson plans to drive 100 miles

Hudson has at most $70

solution

let x represent he maximum number of days Hudson can afford to rent while staying on budget. cost of number of days is:

$31.91 * x= 31.91x

As Hudson plans to ride 100 miles; cost is:

100 * 0.07 = $7

therefore number of days plus 100 miles ride is:

31.91x + 7

with a limit of less than or equal to the budget we have:

$31.91x + 7 \leq 70$

The inequality above could be used to determine x

Read more on inequality here: brainly.com/question/11613554

#SPJ1

6 0

1 year ago

Pls solve the simultaneous equation in the attachment.

siniylev [52]

Answer:

Part a) The solution is the ordered pair (6,10)

Part b) The solutions are the ordered pairs (7,3) and (15,1.4)

Step-by-step explanation:

Part a) we have

$\frac{x}{2}-\frac{y}{5}=1$ ----> equation A

$y-\frac{x}{3}=8$ ----> equation B

Multiply equation A by 10 both sides to remove the fractions

$5x-2y=10$ ----> equation C

isolate the variable y in equation B

$y=\frac{x}{3}+8$ ----> equation D

we have the system of equations

$5x-2y=10$ ----> equation C

$y=\frac{x}{3}+8$ ----> equation D

Solve the system by substitution

substitute equation D in equation C

$5x-2(\frac{x}{3}+8)=10$

solve for x

$5x-\frac{2x}{3}-16=10$

Multiply by 3 both sides

$15x-2x-48=30$

$15x-2x=48+30$

Combine like terms

$13x=78$

$x=6$

<em>Find the value of y</em>

$y=\frac{x}{3}+8$

$y=\frac{6}{3}+8$

$y=10$

The solution is the ordered pair (6,10)

Part b) we have

$xy=21$ ---> equation A

$x+5y=22$ ----> equation B

isolate the variable x in the equation B

$x=22-5y$ ----> equation C

substitute equation C in equation A

$(22-5y)y=21$

solve for y

$22y-5y^2=21$

$5y^2-22y+21=0$

Solve the quadratic equation by graphing

The solutions are y=1.4, y=3

see the attached figure

<em>Find the values of x</em>

For y=1.4

$x=22-5(1.4)=15$

For y=3

$x=22-5(3)=7$

therefore

The solutions are the ordered pairs (7,3) and (15,1.4)

3 0

4 years ago

How to convert 80 kilometers per hour is 60 miles per hour?

Tatiana [17]

1 km = 0.62 mi
1 mi = 1.6 km

80*0.62
-or-
60*1.6

Hope this helps

5 0

3 years ago