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VikaD [51]
3 years ago
14

Figure EFGH on the grid below represents a trapezoidal plate at its starting position on a rotating platform:

Mathematics
1 answer:
Murrr4er [49]3 years ago
4 0

Answer: A is correct

Step-by-step explanation: I took the test and I got 100

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The point (-2, - 5) is on the line given by which equation below?​
Rina8888 [55]

Answer:

The answer to your question is the first choice

Step-by-step explanation:

Data

Point = (-2, -5)

Process

Evaluate each of the equations on the point given.

A. y = x - 3

    -5 = -2 - 3

    -5 = -5     This is true so the point is on this line

B. y = 2x

    -5 = 2(-2)

   -5 = -4       The result is false so the point is not on this line

C. y = -3x

   -5 = -3(-2)

   -5 = 6         The result is false so the point is not on this line

D. y = x + 3

  -5 = -2 + 3

  -5 = 1           The result is false so the point is not on this line

8 0
3 years ago
Read 2 more answers
True or False, 15%-54% of errors reported from primary care practices are related to the testing process.
vovikov84 [41]

Answer:

i would say true

Step-by-step explanation:

5 0
3 years ago
Gymnast Clothing manufactures expensive soccer cleats for sale to college bookstores in runs of up to 500. Its cost (in dollars)
Molodets [167]
What gymnastics with soccer
7 0
3 years ago
The value of the digit 8 in this number is 8 the value of the digit 7 in this number is not 700
In-s [12.5K]
It got to be 800 or something
3 0
3 years ago
The lengths of pregnancies are normally distributed with a mean of 266 days and a standard deviation of 15 days.
inessss [21]

The lengths of pregnancies are normally distributed with a mean of 266 days and a standard deviation of 15 days.

That is,

Consider X be the length of the pregnancy

Mean and standard deviation of the length of the pregnancy.

Mean \mu =266\\

Standard deviation \sigma =15

For part (a) , to find the probability of a pregnancy lasting 308 days or longer:

That is, to find P(X\geq 308)

Using normal distribution,

z=\frac{X-\mu}{\sigma}

z=\frac{308-266}{15}

=\frac{42}{15}

Thus z=2.8

So P\left (X\geq 308  \right )=1-P(X

=1-P(z

=1-Table\:  value\:  of\:  2.8

=1-0.99744

=0.00256

Thus the probability of a pregnancy lasting 308 days or longer is given by 0.00256.

This the answer for part(a): 0.00256

For part(b), to find the length that separates premature babies from those who are not premature.

Given that the length of pregnancy is in the lowest 3​%.

The z-value for the lowest of 3% is -1.8808

Then X=\frac{X-\mu}{\sigma}\Rightarrow X=z*\sigma+\mu

This implies X=-1.8808*15+266=237.788

Thus the babies who are born on or before 238 days are considered to be premature.

5 0
3 years ago
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