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sladkih
7 days ago
7

How many different letter permutations, of any length, can be made using the letters M O T T O (For instance, there are 3 possib

le permutations of length 1.)
Mathematics
1 answer:
tester [3.9K]7 days ago
5 0

Answer:

89

Step-by-step explanation:

Given that

2 O, 2 T and 1 M

Now based on this, the following arrangements exist  

1

Three arrangements i.e. {M,T,O}

2

XX or XY

XX in 2C1 = two arrangements i.e. {OO or TT}

XY in 3C2 × 2! = six arrangements

3

XXY or XYZ

XXY in 2C1 × 2C1 × 3! ÷ 2! = twelve arrangements

XYZ in 3C3 × 3! = six arrangements

4

XXYY or XXYZ

XXYY = 4! ÷ (2! × 2!) = six arrangements

XXYZ in 2C1 ×  4! ÷ 2! = twenty four arrangements

5

= 5! ÷ (2! ×2!)

= 120 ÷ 4

= 30

Thus, the overall total is

= 3 + (2 +6)+ (12 +6) + (6 +24) + 30

= 89

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Answer:

At the α = 0.10 level, there is no substantial evidence indicating that the average vertical jump for students at this school differs from 15 inches.

Step-by-step explanation:

A hypothesis test is necessary to verify the assertion that the average vertical jump of students diverges from 15 inches.

The null and alternative hypotheses are:

H_0: \mu=15\\\\H_a: \mu\neq15

The significance level is set at 0.10.

The sample mean recorded is 17, and the sample standard deviation is 5.37.

The degrees of freedom are calculated as df=(20-1)=19.

The t-statistic is:

t_{19}=\frac{M-\mu}{s_M/\sqrt{n}} =\frac{17-15}{5.37/\sqrt{20}}=\frac{2}{5.37/4.47} =2/1.20=1.67

The two-tailed P-value corresponding to t=1.67 is P=0.11132.

<pSince this P-value exceeds the significance level, the result is not significant. Therefore, the null hypothesis remains unchallenged.

At the α = 0.10 level, there is no compelling evidence that the average vertical jump of students at this school deviates from 15 inches.

7 0
6 days ago
A truck with a heavy load drove from Boston to New York at 50 mph. After dropping of the load, it returned from New York to Bost
Zina [3917]

Answer:

175/3 or 58.333...

Step-by-step explanation:

This problem can be complex, but I'll simplify it. I apologize for the challenges involved in formatting fractions correctly. To ensure accuracy, I've added several parentheses that may seem unnecessary.

Let d denote the distance between the two cities.

The time it takes can be expressed as d/r, where r represents the speed; thus, from B to NY, it is d/50.

In the same manner, from NY to B, it's d/70.

The average velocity can be determined as 2d divided by the sum of d/50 and d/70, representing the average time.

Next, eliminate d from the equation to reach (d/d)*2/(1/50+1/70).

Now, you can multiply by the Least Common Multiple (LCM) over itself, which is 350/350.

The result after performing this calculation will yield 700/12, equating to 58.333...(This is how you would enter it into your RSM browser) I hope this clarification is useful. My apologies for the delay.

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6 days ago
The perimeter of the school crossing sign is 102 inches what is the length of each side
Leona [4187]
The total perimeter of the school crossing sign measures 102 inches.
To find the length of each side, knowing the number of sides is necessary.
If it has 5 sides, dividing 102 by 5 results in 20.4 inches per side.
If it has 4 sides, dividing 102 by 4 gives a side length of 25.5 inches.
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6 0
14 days ago
A line passes through the point (0, -1) and has a positive slope. Which of these points could that line pass through?
zzz [4035]

Answer:

(12, 3), (-2, -5), (1, 15)

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8 2
9 days ago
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Svet_ta [4341]
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7 days ago
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