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3 years ago

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