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2 years ago

Help please writer the answer as letters

Mathematics

2 answers:

xz_007 [3.2K]2 years ago

7 0

the correct answer for A is choice B. answer choice C is correct for question B.

Step-by-step explanation:

answer choice D is correct for C.

Liula [17]2 years ago

3 0

Answer:

B, D, D

Step-by-step explanation:

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Tell which measure is greater. One yard or one meter?

tensa zangetsu [6.8K]

Answer:

one yard one yard is bigger. 1 yard=0.9144 meter

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1 year ago

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Simplify: (14y+x) +22y=??

PtichkaEL [24]

Answer:

x+36y

Step-by-step explanation:

i did it and got a 100%

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1 year ago

Please help me this question

postnew [5]

Let's split it into rectange and tringle
BC= 2.1+ sqrt(3.2^2 - 1.9^2) = 4.67

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2 years ago

Read 2 more answers

5×=15+10y,3×+7y=31 it's for tomorrow pls

Katarina [22]

Answer:

x = 6.39, y = 1.69 is the solution of the given equation system.

Step-by-step explanation:

Here, the given set of equation is:

5 x = 15 + 10 y ⇒ 5 x - 10 y = 15 ...... (1)

3 x +7 y = 31 ............. (2)

Now, the coefficient of x in first equation id 5 and in second is 3.

So, MULTIPLY (1) with 3 and (2) with -5 , we get:

15 x - 30 y = 45

-15 x- 35 y = -155

<u>ADD BOTH EQUATIONS ,</u> we get:

15 x - 30 y -15 x- 35 y = 45 - 155

or, -65 y = -110

or, y = 110/65 = 1.69

Put y = 1.69 in 3 x +7 y = 31 , we get:

3(x) + 7 (1.69) =31

⇒ 3 x = 19.17 or x = 6.39

Hence, x = 6.39, y = 1.69 is the solution of the given equation system

3 0

2 years ago

[Q71 Suppose that the height, in inches, of a 25-year-old man is a normal random variable with parameters g = 71 inch and 02 = 6

viktelen [127]

Answer: (a) Percentage of 25 year old men that are above 6 feet 2 inches is 11.5%.

(b) Percentage of 25 year old men in the 6 footer club that are above 6 feet 5 inches are 2.4%.

Step-by-step explanation:

Given that,

Height (in inches) of a 25 year old man is a normal random variable with mean $g=71$ and variance $o^{2} =6.25$ .

To find: (a) What percentage of 25 year old men are 6 feet, 2 inches tall

(b) What percentage of 25 year old men in the 6 footer club are over 6 feet. 5 inches.

Now,

(a) To calculate the percentage of men, we have to calculate the probability

P[Height of a 25 year old man is over 6 feet 2 inches]= P[X> $74in$ ]

P[X>74] = P[ $\frac{X-g}{o}$ > $\frac{74-71}{2.5}$ ]

= P[Z > 1.2]

= 1 - P[Z ≤ 1.2]

= 1 - Ф (1.2)

= 1 - 0.8849

= 0.1151

Thus, percentage of 25 year old men that are above 6 feet 2 inches is 11.5%.

(b) P[Height of 25 year old man is above 6 feet 5 inches gives that he is above 6 feet] = P[X, $6ft$ $5in$ - X, $6ft$ ]

P[X > $6ft$ $5in$ I X > $6ft$ ] = P[X > 77 I X > 72]

= $\frac{P[X > 77]}{P[ X > 72]}$

= $\frac{P[\frac{X - g}{o}>\frac{77-71}{2.5}] }{P[\frac{X-g}{o} >\frac{72-71}{2.5}] }$

= $\frac{P[Z >2.4]}{P[Z>0.4]}$

= $\frac{1-P[Z\leq2.4] }{1-P[Z\leq0.4] }$

= $\frac{1-0.9918}{1-0.6554}$

= $\frac{0.0082}{0.3446}$

= 0.024

Thus, Percentage of 25 year old men in the 6 footer club that are above 6 feet 5 inches are 2.4%.

4 0

3 years ago