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user100 [1]
3 years ago
9

Tonysha has two bags. Each bag has three objects in it. The first bag has one dime, one nickel, and one penny. The second bag ha

s one red button, one blue button, and one yellow button. Tonysha picks one object from each bag. How many possible combination of two objects can she pick?
Mathematics
1 answer:
Tom [10]3 years ago
8 0
It's 6 hope this helped
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Solve the inequality and graph the solution N+3<5.
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Answer:

n<2

Step-by-step explanation:

move all terms not containing n to the right side of the inequality

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If -11 + N = -11, then N is the
gladu [14]

Answer:additive inverse

Step-by-step explanation:

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3 years ago
What is (9,+9)0-2 and 158932x738995
Mekhanik [1.2K]

Answer:

117,449,953,338.

Step-by-step explanation:

(9,+9)0-2  =  0 - 2

= -2.

158932 x 738995

=  117,449,953,340

So the answer is

117,449,953,340 - 2

= 117,449,953,338.

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3 years ago
Carry out the following division.​
kifflom [539]

Answer:

1: 8a²

2: 9a²b²c²

Step-by-step explanation:

1: 48a³ ÷ 6a

48÷6 = 8

8a²

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72÷ 8 =9

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4 0
2 years ago
Read 2 more answers
Suppose that the mean water hardness of lakes in Kansas is 425 mg/L and these values tend to follow a normal distribution. A lim
tino4ka555 [31]

Answer:

The mean water hardness of lakes in Kansas is 425 mg/L or greater.

Step-by-step explanation:

We are given the following data set:

346, 496, 352, 378, 315, 420, 485, 446, 479, 422, 494, 289, 436, 516, 615, 491, 360, 385, 500, 558, 381, 303, 434, 562, 496

Formula:

\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}  

where x_i are data points, \bar{x} is the mean and n is the number of observations.  

Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}

Mean =\displaystyle\frac{10959}{25} =438.36

Sum of squares of differences = 175413.76

S.D = \sqrt{\dfrac{175413.76}{24}} = 85.49

Population mean, μ = 425 mg/L

Sample mean, \bar{x} = 438.36

Sample size, n = 25

Alpha, α = 0.05

Sample standard deviation, s = 85.49

First, we design the null and the alternate hypothesis

H_{0}: \mu \geq 425\text{ mg per Litre}\\H_A: \mu < 425\text{ mg per Litre}

We use one-tailed t test to perform this hypothesis.

Formula:

t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }

Putting all the values, we have

t_{stat} = \displaystyle\frac{438.36 - 425}{\frac{85.49}{\sqrt{25}} } = 0.7813

Now, t_{critical} \text{ at 0.05 level of significance, 24 degree of freedom } = -1.7108

Since,                        

The calculated t-statistic is greater than the critical value, we fail to reject the null hypothesis and accept it.

Thus, the mean water hardness of lakes in Kansas is 425 mg/L or greater.

6 0
3 years ago
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