Which statements are correct descriptions of the solid

You can clearly eliminate 0.79d as an answer, because that is less than the amount that Ron's class donated. The last answer is also incorrect, because it states that Danny's class donated 21 more cents than Ron's class. The problem says that Danny's class donated 21% more than Ron's. The only way that d + 0.21 would be correct is if Ron's class donated $1. The problem does not say that. Therefore, we are left with two possibilities. The first choice is the correct one. Why?

21% more than Ron's class' donation would be the amount that Ron's class donated + an additional 21%. If Ron's class donated d dollars, then an additional 21% would be 0.21 * d = 0.21d. Hence, Danny's class donated a total of:

d + 0.21d = d(1 + 0.21) = 1.21d

6 0

3 years ago

Which expressions are equivalent to the given expression? Select two options.

Mars2501 [29]

Answer : The two correct options are,

3.4a + 5.6b

5.6b + 3.4a

Step-by-step explanation :

The given expression is:

(-2.4a - 1.8b) - (-3.8a - 4.4b) + (2.0a + 3.0b)

First we have to open the brackets.

⇒ -2.4a - 1.8b + 3.8a + 4.4b +2.0a + 3.0b

Now keep like terms together.

⇒ -2.4a + 3.8a +2.0a + 3.0b - 1.8b + 4.4b

By subtraction or addition, we get:

⇒ 3.4a + 5.6b

or,

⇒ 5.6b + 3.4a

Hence, the expressions that equivalent to the given expression are, 3.4a + 5.6b and 5.6b + 3.4a

7 0

3 years ago

Read 2 more answers

A process manufactures ball bearings with diameters that are normally distributed with mean 25.1 mm and standard deviation 0.08

marta [7]

Answer:

(a) The proportion of the diameters are less than 25.0 mm is 0.1056.

(b) The 10th percentile of the diameters is 24.99 mm.

(c) The ball bearing that has a diameter of 25.2 mm is at the 84th percentile.

(d) The proportion of the ball bearings meeting the specification is 0.8881.

Step-by-step explanation:

Let X = diameters of ball bearings.

The random variable X is normally distributed with mean, μ = 25.1 mm and standard deviation, σ = 0.08 mm.

To compute the probability of a Normally distributed random variable we need to first convert the raw scores to z-scores as follows:

z = (X - μ) ÷ σ

(a)

Compute the probability of X < 25.0 mm as follows:

P (X < 25.0) = P ((X - μ)/σ < (25.0-25.1)/0.08)

= P (Z < -1.25)

= 1 - P (Z < 1.25)

= 1 - 0.8944

= 0.1056

*Use a z-table for the probability.

Thus, the proportion of the diameters are less than 25.0 mm is 0.1056.

(b)

The 10th percentile implies that, P (X < x) = 0.10.

Compute the 10th percentile of the diameters as follows:

P (X < x) = 0.10

P ((X - μ)/σ < (x-25.1)/0.08) = 0.10

P (Z < z) = 0.10

z = -1.282

The value of x is:

z = (x - 25.1)/0.08

-1.282 = (x - 25.1)/0.08

x = 25.1 - (1.282 × 0.08)

= 24.99744

≈ 24.99

Thus, the 10th percentile of the diameters is 24.99 mm.

(c)

Compute the value of P (X < 25.2) as follows:

P (X < 25.2) = P ((X - μ)/σ < (25.2-25.1)/0.08)

= P (Z < 1.25)

= 0.8944

≈ 0.84

*Use a z-table for the probability.

Thus, the ball bearing that has a diameter of 25.2 mm is at the 84th percentile.

(d)

Compute the value of P (25.0 < X < 25.3) as follows:

P (25.0 < X < 25.3) = P ((25.0-25.1)/0.08 < (X - μ)/σ < (25.3-25.1)/0.08)

= P (-1.25 < Z < 2.50)

= P (Z < 2.50) - P (Z < -1.25)

= 0.99379 - 0.10565

= 0.88814

≈ 0.8881

*Use a z-table for the probability.

Thus, the proportion of the ball bearings meeting the specification is 0.8881.

4 0

3 years ago