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3 years ago

Hi, please help me with this thank you!

Mathematics

1 answer:

sdas [7]3 years ago

5 0

5 hope this helps sorry if wrong

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WINSTONCH [101]

DIVIDE THE 920 BY 7.50 AND YOU SHOULD GET YOUR ANSWER

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3 years ago

Read 2 more answers

Wich measure is less than 435 inches

podryga [215]

435 inches = 36 1/4 ft.

So, the answer would be D) 12 ft. 3 in. hope this helps

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3 years ago

In the midpoint rule for triple integrals we use a triple riemann sum to approximate a triple integral over a box b, where f(x,

Lana71 [14]

The sub-boxes will have dimensions $\frac{2-0}{2} \times \frac{2-0}{2} \times \frac{2-0}{2} =1\times1\times1=1 \ cubic \ units$

x sub-intervals are 0 to 1 and 1 to 2. Midpoints are at $x= \frac{1}{2}$ and $x= \frac{3}{4}$
y sub-intervals are 0 to 1 and 1 to 2. Midpoints are at $y= \frac{1}{2}$ and $y= \frac{3}{4}$
z sub-intervals are 0 to 1 and 1 to 2. Midpoints are at $z= \frac{1}{2}$ and $z= \frac{3}{4}$ 

Let $f(x,y,z)=\cos{(xyz)}$

$\int\limits \int\limits \int\limits {f(x,y,z)} \, dV \approx f\left( \frac{1}{2} , \frac{1}{2} , \frac{1}{2} \right)+f\left( \frac{1}{2} , \frac{1}{2} , \frac{3}{4} \right)+f\left( \frac{1}{2} , \frac{3}{4} , \frac{1}{2} \right)+f\left( \frac{1}{2} , \frac{3}{4} , \frac{3}{4} \right)$
$+f\left( \frac{3}{4} , \frac{1}{2} , \frac{1}{2} \right)+f\left( \frac{3}{4} , \frac{1}{2} , \frac{3}{4} \right)+f\left( \frac{3}{4} , \frac{3}{4} , \frac{1}{2} \right)+f\left( \frac{3}{4} , \frac{3}{4} , \frac{3}{4} \right) \\ \\ \approx\cos{ \frac{1}{8} }+\cos{ \frac{3}{16} }+\cos{ \frac{3}{16} }+\cos{ \frac{9}{32} }+\cos{ \frac{3}{16} }+\cos{ \frac{9}{32} }+\cos{ \frac{9}{32} }+\cos{ \frac{27}{64} } \\ \\ \approx0.9922+0.9825+0.9825+0.9607+0.9825+0.9607+0.9607 \\ +0.9123 \\ \\ \approx\bold{7.734}$

5 0

3 years ago

Question Help Suppose that the lifetimes of light bulbs are approximately normally distributed, with a mean of 5656 hours and a

koban [17]

Answer:

a)3.438% of the light bulbs will last more than 6262 hours.

b)11.31% of the light bulbs will last 5252 hours or less.

c) 23.655% of the light bulbs are going to last between 5858 and 6262 hours.

d) 0.12% of the light bulbs will last 4646 hours or less.

Step-by-step explanation:

Normally distributed problems can be solved by the z-score formula:

On a normaly distributed set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a value X is given by:

$Z = \frac{X - \mu}{\sigma}$

After we find the value of Z, we look into the z-score table and find the equivalent p-value of this score. This is the probability that a score will be LOWER than the value of X.

In this problem, we have that:

The lifetimes of light bulbs are approximately normally distributed, with a mean of 5656 hours and a standard deviation of 333.3 hours.

So $\mu = 5656, \sigma = 333.3$

(a) What proportion of light bulbs will last more than 6262 hours?

The pvalue of the z-score of $X = 6262$ is the proportion of light bulbs that will last less than 6262. Subtracting 100% by this value, we find the proportion of light bulbs that will last more than 6262 hours.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{6262 - 5656}{333.3}$

$Z = 1.82$

$Z = 1.81$ has a pvalue of .96562. This means that 96.562% of the light bulbs are going to last less than 6262 hours. So

$P = 100% - 96.562% = 3.438%$ of the light bulbs will last more than 6262 hours.

(b) What proportion of light bulbs will last 5252 hours or less?

This is the pvalue of the zscore of $X = 5252$

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{5252- 5656}{333.3}$

$Z = -1.21$

$Z = -1.21$ has a pvalue of .1131. This means that 11.31% of the light bulbs will last 5252 hours or less.

(c) What proportion of light bulbs will last between 5858 and 6262 hours?

This is the pvalue of the zscore of $X = 6262$ subtracted by the pvalue of the zscore $X = 5858$

For $X = 6262$ , we have that $Z = 1.81$ with a pvalue of .96562.

For X = 5858

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{5858- 5656}{333.3}$

$Z = 0.61$

$Z = 0.61$ has a pvalue of .72907.

So, the proportion of light bulbs that will last between 5858 and 6262 hours is

$P = .96562 - .72907 = .23655$

23.655% of the light bulbs are going to last between 5858 and 6262 hours.

(d) What is the probability that a randomly selected light bulb lasts less than 4646 hours?

This is the pvalue of the zscore of $X = 4646$

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{4646- 5656}{333.3}$

$Z = -3.03$

$Z = -3.03$ has a pvalue of .0012. This means that 0.12% of the light bulbs will last 4646 hours or less.

5 0

3 years ago

ASAP PLS HELP DUE TODAY. WILL GIVE BRAINLIEST IF A GOOD ANSWER EXPLENATION AND CORRECT ANSWER

aniked [119]

Answer:

$7 + (-8)$

Step-by-step explanation:

Given

$7 - 8$

(a): Write as additive inverse.

An additive inverse is of the form a + (-b)

In this case:

$a = 7$

$b = -8$

So, the expression can be represented as:

$7 + (-8)$

(b): Number line representation

When the expression in (a) is solved.

The result is:

$7 - 8 = 7 + (-8) = -1$

This means that the number line must accommodate 7 and -1.

Having said that, options (b) and (c) are out because their range is 0 to 15 and this excludes -1.

Option (d) is a wrong representation of $7 - 8$

Hence, (a) is correct

7 0

3 years ago