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Andreas93 [3]
4 years ago
6

Baseball Ichiro Suzuki holds the American League record for the most hits in a single baseball season. In 2004, Suzuki had a tot

al of 262 hits for the Seattle Mariners. He hit three fewer triples than home runs, and he hit three times as many doubles as home runs. Suzuki also hit 45 times as many singles as triples. Find the number of singles, doubles, triples, and home runs hit by Suzuki during the season.
Mathematics
1 answer:
mamaluj [8]4 years ago
7 0

Let Home runs = X

Triples would be X-3 ( 3 less triples than home runs)

Doubles would be 3x ( 3 times as many doubles as home runs)

Singles would be 45(x-3) ( 45 times as many singles as triples)

Simplify the equation for singles to be 45x-153

Now you have X + x-3 + 3x + 4x-135 = 262

Simplify:

50x - 138 = 262

Add 138 to both sides:

50x = 400

Divide both sides by 50:

x = 400/50

x = 8

Home runs = x = 8

Triples = x-3 = 8-3 = 5

Doubles = 3x = 3(8) = 24

Singles = 45(x-3) = 45(8-3) = 45(5) = 225

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A small shipping crate can hold up to 500 pounds. Kelly needs to ship two machine parts that are 255 pounds and 321 pounds. She
Olenka [21]

Answer: no they don't both fit in one box.

Step-by-step explanation: first off, 200 plus 300 is 500, and both numbers are bigger. Then if you wanted to round it to the nearest 10, you would get 580. 580>500 so yeah. that won't work.

6 0
2 years ago
A production facility employs 10 workers on the day shift, 8 workers on the swing shift, and 6 workers on the graveyard shift. A
PilotLPTM [1.2K]

Answer:

252 ; 0.0059 ; 0.0074 ; 0.9926

Step-by-step explanation:

Day shift = 10

Swing shift = 8

Graveyard shift = 6

Total number of workers = 24

A.) Number of selections resulting in 5 workers coming from day shift :

10C5 = 10! ÷ (10-5)!5!

= (10*9*8*7*6) / (5*4*3*2*1)

= 252

B.) All 5 workers coming from day shift :

Required outcome = 10C5

Total possible outcomes = 24C5

10C5 ÷ 24C5

252 ÷ 42504

= 0.0059288

= 0.0059

C.) 5 selected workers are from the same shift :

[day shift + swing shift + graveyard shift] / total possible outcomes

[(10C5) + (8C5) + (6C5)] ÷ 24C5

(252 + 56 + 6) / 42504

= 0.0074

D.) What is the probability that at least two different shifts will be represented among the selected workers?

1 - [[(10C5) + (8C5) + (6C5)] ÷ 24C5]

1 - 0.0073875

= 0.9926124

= 0.9926

6 0
3 years ago
X&gt;y <br><br>Which is greater? x+a or y+a urgent please?!
earnstyle [38]

Answer:

x+a .....................

4 0
4 years ago
Read 2 more answers
If anyone knows this I will give them brainiest
statuscvo [17]

Answer:

x = 12

m(QS) = 52°

m(PD) = 152°

Step-by-step explanation:

Recall: Angle formed by two secants outside a circle = ½(the difference of the intercepted arcs)

Thus:

m<R = ½[m(PD) - m(QS)]

50° = ½[(12x + 8) - (4x + 4)] => substitution

Solve for x

Multiply both sides by 2

2*50 = (12x + 8) - (4x + 4)

100 = (12x + 8) - (4x + 4)

100 = 12x + 8 - 4x - 4 (distributive property)

Add like terms

100 = 8x + 4

100 - 4 = 8x

96 = 8x

96/8 = x

12 = x

x = 12

✔️m(QS) = 4x + 4 = 4(12) + 4 = 52°

✔️m(PD) = 12x + 8 = 12(12) + 8 = 152°

3 0
3 years ago
The top and bottom margins of a poster are each $3$ cm and the side margins are each $2$ cm. If the area of printed material on
babymother [125]

Answer:

Therefore the dimension of the poster is 12 cm by 8 cm.

Step-by-step explanation:

Let the length of the poster be x and the width be y.

Given that the area of the poster is 96 cm².

∴xy =96

\Rightarrow y= \frac{96}{x}

The sides margins each are 2 cm and the top and bottom margins of the poster are each 3 cm.

The length of printing space is =(x- 2.3) cm

                                                   = (x-6) cm

The width of the printing space is =(y-2.2) cm

                                                         =( y-4 )cm

The area of the printing space is A=(x-6)(y-4) cm²

∴A=(x-6)(y-4)  

\Rightarrow A=(x-6)(\frac{96}{x}-4)    [ Putting y= \frac{96}{x}  ]

\Rightarrow A=120-\frac{576}{x}-4x

Differentiating with respect to x

A'= \frac{576}{x^2}-4

Again differentiating with respect to x

A''=-\frac{1152}{x^3}

To find the minimum area, we set A'=0

\therefore  \frac{576}{x^2}-4=0

\Rightarrow \frac{576}{x^2}=4

\Rightarrow x^2=\frac{576}{4}

\Rightarrow x^2 =144

\Rightarrow x=\pm 12

Dimension can't be negative.

Therefore x=12

If x=12, the value of A''>0,then at x=12, the area of the poster will be minimum.

If x=12, the value of A''<0,then at x=12, the area of the poster will be minimum.

\therefore A''|_{x=12}=-\frac{1152}{12^3}

Therefore at x= 12 cm the area of the poster will be maximum.

The width of the poster is y=\frac{96}{12} = 8 cm

Therefore the dimension of the poster is 12 cm by 8 cm.

3 0
3 years ago
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