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3 years ago

The question is on the sheet

Mathematics

1 answer:

Ghella [55]3 years ago

6 0

Y=1/2x-5 is the answer

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A soccer team scores an average of 11 goals per season. A double number line is drawn to figure out the average

Allushta [10]

Answer:

it c

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

What is the length of the hypotenuse? Answer : 17

leonid [27]

Answer:

Below.

Step-by-step explanation:

Finding the hypotenuse:

Recall the formula of Pythagoras'theorem,

(Hypotenuse)² = (Base)² + (Perpendicular)²

Given that base = 8 and perpendicular = 15

Plugging the values,we obtain

H² = 8² + (15)²
H² = 64 + 225
H² = 289
H = √(17*17)
H = 17

Now applying trigonometric functions:

Recall SOHCAHTOA , it's very helpful.

Sin θ = Sin A = Opposite/Hypotenuse

Sin A = 15/17 ≈ 0.88

cos θ = cos A = Adjacent/Hypotenuse

cos A = 8/17 ≈ 0.47

Tan θ = tan A = Opposite/Adjacent

tan A = 15/8 ≈ 1.88

Note: The values in decimal are rounded to nearest hundredth.

3 0

2 years ago

What is m, 2/3 × 4/5 = 1/3÷ m

ankoles [38]

Answer:

$\frac{2}{3} \times \frac{4}{5} = \frac{1}{3} \div m \\ \frac{8}{15} = \frac{1}{3m} \\ \frac{8}{5} = \frac{1}{m} \\ 8m = 5 \\ m = \frac{5}{8} \\$

4 0

4 years ago

Read 2 more answers

A certain large manufacturing facility produces 20,000 parts each week. The manager of the facility estimates that about 1% of t

Slav-nsk [51]

Answer:

198 is the variance for the number of defective parts made each week.

Step-by-step explanation:

We are given the following in the question:

Number of parts produced each week = 20,000

Percentage of defective parts = 1%

We have to calculate the variance for the number of defective parts made each week.

We treat defective part as a success.

P(Defective part) = 1% = 0.01

$p = 0.01$

Then the number defective parts follows a binomial distribution .

Formula for variance =

$\sigma^2 = np(1-p) = 20000(0.01)(1-0.01) = 198$

Thus, 198 is the variance for the number of defective parts made each week.

8 0

4 years ago

John read the first 1 1 4 114114 pages of a novel, which was 3 33 pages less than 1 3 3 1 ? start fraction, 1, divided by, 3, en

damaskus [11]

We are given statement " John read the first 114 pages of a novel, which was 3 pages less than 1/3 of the novel".

Let us assume total number of pages of a novel = p pages.

3 less than 1/3 of total pages = 114.

3 less then 1/3 of P is = 114.

$\frac{1}{3}P- 3 = 114$

<h3>Therefore, required equation for number of pages of a novel is </h3><h3> $\frac{1}{3}P- 3 = 114$ .</h3>

8 0

3 years ago

The question is on the sheet ​

The question is on the sheet