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1 year ago

MUSIC HELP plss ahh pls

SAT

1 answer:

Mashutka [201]1 year ago

4 0

Answer:

THERE'S A 15 NOTE COUNTS

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For the month of in a certain city, the probability that the weather on a given day is is. Also in the month of in the same ci

7nadin3 [17]

Answer:

Wednesday summer

Explanation:

summer day

8 0

2 years ago

A 100-n weight is hung at the middle of a 4-m long horizontal clothesline. If the middle of the clothesline sags by 1 m, what is

Elena L [17]

The tension on each segment of the clothesline is : 110 N

<u>Given data : </u>

mass of object = 100 n = 10 kg

Horizontal distance of clothesline = 4 m

middle of clothesline sag s by : 1m

<h3 /><h3>Determine the tension on each segment of clothesline</h3>

<u>First step</u> : calculate the horizontal angle made by the sagging

β = arctan ( 1 m / 2m ) ----- ( 1 )

= arctan ( 0.5 )

≈ 26.57°

Note : Tension in th y axis ( Ty ) = Tsinβ

Therefore :

Tension on each segment can be calculated using the formula below

2Tsinβ - mg = 0

solve for T

T = mg / 2sinβ

= ( 10 * 9.8 ) / 2 * sin 26.57°

= 98 / 0.89

= 110 N

Hence we can conclude that the tension on each segment of the clothesline is : 110 N

Learn more about Tension calculations : brainly.com/question/24994188

7 0

2 years ago

the distribution of lengths of salmon from a certain river is approximately normal with standard deviation 3.5 inches. if 10 per

juin [17]

Complete question;

<em>The distribution of lengths of salmon from a certain river is approximately normal with a standard deviation of 3.5 inches. If 10 percent of salmon are longer than 30 inches, which of the following is closest to the mean of the distribution? 26 inches A 28 inches B 30 inches C 33 inches D 34 inches</em>

Option B is correct. The value that is closest to the mean of the distribution is 30inches.

The formula for calculating the z-score is expressed as:

$z=\frac{ x-\mu}{\sigma}$

Given the following parameters

$\sigma = 3.5\\x = 26in$

If 10 percent of salmon are longer than 30 inches, then:

P(x>30) = 0.1

Using the z table to get the value corresponding to the mean area 0.1.

The required z-score will be -1.285

Substitute the resulting parameters into the formula to get the mean of the distribution.

$z=\frac{x-\mu}{\sigma}\\-1.285=\frac{26-\mu}{3.5}\\-1.285 \times 3.5 = 26 - \mu\\ -4.4975 = 26 - \mu\\\mu = 26 + 4.4975\\\mu = 30.4975\\\mu \approx 30in$