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2 years ago

How many fewer people attended the pacers games than atteneded the clippers game

2 answers:

6 0

Do you have numbers for each game. If so subtract the bigger number from higher then your answer is there.

say you have 189 people at the pacer game and 123 at the clippers you take 189-123=66
so 66 fewer people attened

diamong [38]2 years ago

5 0

so 66 fewer people attened

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In a survey of 12 people, it was found that an average of $36 was spent on their child's last birthday gift with a standard devi

Nataliya [291]

Answer:

(27.3692 ; 44.6308)

Step-by-step explanation:

Mean, xbar = 36

Standard deviation, s = 11

Sample size, n = 12

Tcritical at 0.2, df = 12 - 1 = 11 ; Tcritical = 2.718

Confidence interval :

Xbar ± Margin of error

Margin of Error = Tcritical * s/sqrt(n)

Margin of Error = 2.718 * 11/sqrt(12) = 8.6308

Confidence interval :

Lower boundary : 36 - 8.6308 = 27.3692

Upper boundary : 36 + 8.6308 = 44.6308

(27.3692 ; 44.6308)

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2 years ago

Jill says that 12.6 is less than 12.63. explain her error

kenny6666 [7]

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2 years ago

Jay is 3 years less than 4 times Nelly’s age ( n ). Which expression represents Jay’s age?

NeX [460]

Answer:

Step-by-step explanation: 3x4= 12 -3=9

6 0

2 years ago

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In Joe's fish tank, there are 5 goldfish and 9 snails. How does the ratio 9:14 describe Joe's fish tank?

Natalija [7]

The answer is C, number of snails to the total number of animals

4 0

2 years ago

Es 18 de junio de 1815, las tropas napoleónicas se encuentran justo adelante, tu eres el ingeniero en balística y el general Wel

Airida [17]

Answer:

89.1° or -1.4°

Step-by-step explanation:

1. Location:

You are on the Mont-Saint-Jean escarpment, near the Belgian town of Waterloo.

The French troops are about 50 m below you and 1.2 km distant.

2. Finding the firing angle

Data:

R = 1200 m

u = 600 m/s

h = -50 m (the height of the target)

a = 9.8 m/s²

We have two conditions.

Horizontal distance

(1) 1200 = 600t cosθ

Vertical distance

(2) -50 = 600t sinθ - 4.9t²

Divide each side of (1) by 600cosθ.

$(3) \, t =\dfrac{2}{\cos \theta}$

Substitute (3) into (2)

$-50 = 600t \sin \theta - 4.9t^{2} = 600 \left( \dfrac{2}{\cos \theta} \right ) \sin \theta - 4.9 \left( \dfrac{2}{\cos \theta} \right )^{2}\\\\(4) \, -50 = 1200 \tan \theta - \dfrac{19.6}{\cos^{2} \theta}$

Recall that

(5) sec²θ = 1/cos²θ = tan²θ + 1

Substitute (5) into (4)

$-50 = 1200 \tan \theta - 19.6 \left(\tan^{2} \theta}+ 1\right )$

Set up a quadratic equation

$\begin{array}{rcl}-50 & = & 1200 \tan \theta - 19.6\tan^{2} \theta -19.6 \\0 & = & 1200 \tan \theta - 19.6\tan^{2} \theta + 30.4\\0 & =&19.6\tan^{2} \theta - 1200 \tan \theta - 30.4\\0 & =&\tan^{2} \theta - 61.224 \tan \theta - 1.551\\\end{array}$

Solve for θ

Use the quadratic formula.

tanθ = 61.249 or -0.025

θ = arctan(61.249) = 89.1° or

θ = arctan(-0.025) = -1.4°

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2 years ago