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3 years ago

5(x+5)-25=6(2-x) can someone help me with this sum please including how you worked it out

Mathematics

1 answer:

alekssr [168]3 years ago

5 0

Answer:

The answer is 12/11

Step-by-step explanation:

5x+25-25=12-6x

5x=12-6x

11x=12

x=12/11

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Use the area of the rectangle to find the area of the triangle

bezimeni [28]

Answer:

The area of the Triangle is 1/2 the area of the Rectangle

The answer is D.

Step-by-step explanation:

Ignore the area of the rectangle for the moment, just so you can get the answer.

Area Triangle = 1/2 * b * h

Area Triangle = 1/2 * 8 * 4

Area Triangle = 1/2 32

Area Triangle = 16 square feet.

The area of the rectangle = L * W

L = 8

W = 4

Area of the rectangle = 8*4 = 32

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3 years ago

Find the length of arc JL if the radius = 30 cm to the nearest hundredth.

erma4kov [3.2K]

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4.13

Step-by-step explanation:

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2 years ago

Rama jumps 2m 30cm long and swathi jumps 2m 60cm in a competition what is the total distance they jumped together

Furkat [3]

Answer:

Step-by-step explanation:

2 m 30 cm

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2 years ago

The angle of elevation from

weqwewe [10]

Answer:

104.5 ft

Step-by-step explanation:

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5 0

3 years ago

The length of human pregnancies from conception to birth follows a distribution with mean 266 days and standard deviation 15 day

Katena32 [7]

Answer:

a) 81.5%

b) 95%

c) 75%

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 266 days

Standard Deviation, σ = 15 days

We are given that the distribution of length of human pregnancies is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(between 236 and 281 days)

$P(236 \leq x \leq 281)\\\\= P(\displaystyle\frac{236 - 266}{15} \leq z \leq \displaystyle\frac{281-266}{15})\\\\= P(-2 \leq z \leq 1)\\\\= P(z \leq 1) - P(z < -2)\\= 0.838 - 0.023 = 0.815 = 81.5\%$

b) a) P(last between 236 and 296)

$P(236 \leq x \leq 281)\\\\= P(\displaystyle\frac{236 - 266}{15} \leq z \leq \displaystyle\frac{296-266}{15})\\\\= P(-2 \leq z \leq 2)\\\\= P(z \leq 2) - P(z < -2)\\= 0.973 - 0.023 = 0.95 = 95\%$

c) If the data is not normally distributed.

Then, according to Chebyshev's theorem, at least $1-\dfrac{1}{k^2}$ data lies within k standard deviation of mean.

For k = 2

$1-\dfrac{1}{(2)^2} = 75\%$

Atleast 75% of data lies within two standard deviation for a non normal data.

Thus, atleast 75% of pregnancies last between 236 and 296 days approximately.

7 0

3 years ago