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34kurt
3 years ago
9

In ΔKLM, l = 820 cm, m = 560 cm and ∠K=169°. Find the length of k, to the nearest centimeter.

Mathematics
2 answers:
worty [1.4K]3 years ago
8 0

Answer:

1374

Step-by-step explanation:

delta math

Darya [45]3 years ago
6 0

Answer:

1374

Step-by-step explanation:k=

1887526.40528 ≈1373.873≈1374

​

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The diameter of the base of a cone measures 8 units. The height of the cone is 6 units. What is the volume of the cone in terms
FrozenT [24]

Answer:

32pi units3

Step-by-step explanation:

100.53≈32pi

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3 years ago
Jakes has $20 to spend on notebooks and pencils the notebooks and pencils. the notebooks cost 3.25 and the pencils cost $.50 wha
Leokris [45]
Hey there. So basically, find out how much the pencils and notebooks cost first.
The notebooks cost = $3.25
The pencils cost = $0.50

Then, think about what you need to figure out in this problem.
Jake has $20. You need to find how many notebooks Jake can buy in maximum after buying 8 pencils.

If Jake buys 8 pencils that costs $0.50 each, he spends $4 on the pencils.

So now, to find out how many notebooks he can buy, do 20 minus 4.
Jake's got $16 left.

If the notebooks cost $3.25 each, we need to find out how many notebooks he can buy by dividing them. So, 16 divided by 3.25 equals 4.923... and so on.

That means, Jake can buy 4 notebooks with his remaining money.
3 0
3 years ago
No links please :(
kotegsom [21]
A.


Store A has sold the most wristbands
5 0
3 years ago
Suppose it is known that the distribution of purchase amounts by customers entering a popular retail store is approximately norm
iragen [17]

Answer:

a. 0.691

b. 0.382

c. 0.933

d. $88.490

e. $58.168

f. 5th percentile: $42.103

95th percentile: $107.897

Step-by-step explanation:

We have, for the purchase amounts by customers, a normal distribution with mean $75 and standard deviation of $20.

a. This can be calculated using the z-score:

z=\dfrac{X-\mu}{\sigma}=\dfrac{85-75}{20}=\dfrac{10}{20}=0.5\\\\\\P(X

The probability that a randomly selected customer spends less than $85 at this store is 0.691.

b. We have to calculate the z-scores for both values:

z_1=\dfrac{X_1-\mu}{\sigma}=\dfrac{65-75}{20}=\dfrac{-10}{20}=-0.5\\\\\\z_2=\dfrac{X_2-\mu}{\sigma}=\dfrac{85-75}{20}=\dfrac{10}{20}=0.5\\\\\\\\P(65

The probability that a randomly selected customer spends between $65 and $85 at this store is 0.382.

c. We recalculate the z-score for X=45.

z=\dfrac{X-\mu}{\sigma}=\dfrac{45-75}{20}=\dfrac{-30}{20}=-1.5\\\\\\P(X>45)=P(z>-1.5)=0.933

The probability that a randomly selected customer spends more than $45 at this store is 0.933.

d. In this case, first we have to calculate the z-score that satisfies P(z<z*)=0.75, and then calculate the X* that corresponds to that z-score z*.

Looking in a standard normal distribution table, we have that:

P(z

Then, we can calculate X as:

X^*=\mu+z^*\cdot\sigma=75+0.67449\cdot 20=75+13.4898=88.490

75% of the customers will not spend more than $88.49.

e. In this case, first we have to calculate the z-score that satisfies P(z>z*)=0.8, and then calculate the X* that corresponds to that z-score z*.

Looking in a standard normal distribution table, we have that:

P(z>-0.84162)=0.80

Then, we can calculate X as:

X^*=\mu+z^*\cdot\sigma=75+(-0.84162)\cdot 20=75-16.8324=58.168

80% of the customers will spend more than $58.17.

f. We have to calculate the two points that are equidistant from the mean such that 90% of all customer purchases are between these values.

In terms of the z-score, we can express this as:

P(|z|

The value for z* is ±1.64485.

We can now calculate the values for X as:

X_1=\mu+z_1\cdot\sigma=75+(-1.64485)\cdot 20=75-32.897=42.103\\\\\\X_2=\mu+z_2\cdot\sigma=75+1.64485\cdot 20=75+32.897=107.897

5th percentile: $42.103

95th percentile: $107.897

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3 years ago
"Tim's lawn is 20 meters by 30 meters. If Tim can mow 8 square meters of grass per minute, how long will it take him to mow the
frosja888 [35]
One hour and a half
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3 years ago
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