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3 years ago

What would you round 39.995 to the nearest cent?

Mathematics

2 answers:

weqwewe [10]3 years ago

5 0

40.00 because the 5 rounds the 9s to 0s

Lostsunrise [7]3 years ago

4 0

To round the number to nearest cents, we need our answer to be in the hundredths place. From the given, the number in the hundredths place is 9 and after it is 5. We need to round up the number such that the near number becomes 40.00.

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Point B has coordinates (4,1). The x-coordinate of point A is -4. The distance between point A and point B is 10 units. What a

klasskru [66]

Given:

Point B has coordinates (4,1).

The x-coordinate of point A is -4.

The distance between point A and point B is 10 units.

To find:

The possible coordinates of point A.

Solution:

Let the y-coordinate of point A be y. Then the two points are A(-4,y) and B(4,1).

Distance formula:

$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

The distance between point A and point B is 10 units.

$\sqrt{(4-(-4))^2+(1-y)^2}=10$

Taking square on both sides, we get

$(8)^2+(1-y)^2=100$

$(1-y)^2=100-64$

$(1-y)^2=36$

Taking square root on both sides, we get

$(1-y)=\pm \sqrt{36}$

$-y=\pm 6-1$

$y=1\mp 6$

$y=1-6$ and $y=1+6$

$y=-5$ and $y=7$

Therefore, the possible coordinates of point A are either (-4,-5) or (-4,7).

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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3 years ago

A manufacturer of Christmas light bulbs knows that 10% of these bulbs are defective. It is known that light bulbs are defective

Inessa05 [86]

Answer:

(a) P(X $\leq$ 20) = 0.9319

(b) Expected number of defective light bulbs = 15

Step-by-step explanation:

We are given that a manufacturer of Christmas light bulbs knows that 10% of these bulbs are defective. It is known that light bulbs are defective independently. A box of 150 bulbs is selected at random.

Firstly, the above situation can be represented through binomial distribution, i.e.;

$P(X=r) = \binom{n}{r} p^{r} (1-p)^{2} ;x=0,1,2,3,....$

where, n = number of samples taken = 150

r = number of success

p = probability of success which in our question is % of bulbs that

are defective, i.e. 10%

Now, we can't calculate the required probability using binomial distribution because here n is very large(n > 30), so we will convert this distribution into normal distribution using continuity correction.

So, Let X = No. of defective bulbs in a box

Mean of X, $\mu$ = $n \times p$ = $150 \times 0.10$ = 15

Standard deviation of X, $\sigma$ = $\sqrt{np(1-p)}$ = $\sqrt{150 \times 0.10 \times (1-0.10)}$ = 3.7

So, X ~ N( $\mu = 15, \sigma^{2} = 3.7^{2})$

Now, the z score probability distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

(a) Probability that this box will contain at most 20 defective light bulbs is given by = P(X $\leq$ 20) = P(X < 20.5) ---- using continuity correction

P(X < 20.5) = P( $\frac{X-\mu}{\sigma}$ < $\frac{20.5-15}{3.7}$ ) = P(Z < 1.49) = 0.9319

(b) Expected number of defective light bulbs found in such boxes, on average is given by = E(X) = $n \times p$ = $150 \times 0.10$ = 15.

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3 years ago