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3 years ago

Plz answer. I’m in a hurry thx.

Mathematics

1 answer:

antoniya [11.8K]3 years ago

5 0

Answer:

A or D

Step-by-step explanation:

I saw your previous post and saw all of the list and compared it!

I don't think you will type it out in numbers without square root, so that will leave with an option of A,D,E.

I think A is too less, so maybe D?

well which is possible?

A or D

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Sin theta=-1/2 then cos theta=

zepelin [54]

Answer:

Step-by-step explanation:

Step one:

Applying the SOH CAH TOA principle

assuming all dimensions are in cm

Step two:

Given data

opposite= 1 cm

hypotenuse= 2 cm

<h3>we can now solve for θ</h3>

Sin(θ)= opp/hyp

Sin(θ)= 1/2

Sin(θ)= 0.5

θ= sin-1 0.5

θ= 30°

hence from tables cos 30°= 0.866

4 0

3 years ago

A rectangle has a length that is 2 units longer than the width. If the width is increased by 4 units and length increased by 3 u

svp [43]

N is the width and s is the length You can figure the rest out. Ther are many ways to do this. I believe in you.

5 0

3 years ago

A lot of 25 skylight covers are received at your construction site, and before installation are subjected to an acceptance testi

8090 [49]

Answer:

If the lot has 4 defective covers out of 25 total covers, the probability of accepting the lot is P=0.98.

Step-by-step explanation:

We have a population of N=25 skylight covers, were K=4 are defective.

We sample n=5 covers, and we will accept the lot if k=2 or fewer are defective.

We will use the hypergeometric distribution to model this probabilities.

First, to be accepted, the sample can have 2, 1 or 0 defective covers, so the probability of being accepted is:

$P(accepted)=P(k\leq2)=P(k=0)+P(k=1)+P(k=2)$

The probability that there are k defective covers in the sample is:

$P(k)=\dfrac{\dbinom{k}{k}\dbinom{N-k}{n-k}}{\dbinom{N}{n}}$

Then, we can calculate the individual probabilities as:

$P(k=0)=\dfrac{\dbinom{4}{0}\cdot \dbinom{25-4}{5-0}}{\dbinom{25}{5}}=\dfrac{\dbinom{4}{0}\cdot \dbinom{21}{5}}{\dbinom{25}{5}}\\\\\\P(k=0)=\dfrac{1\cdot 20349}{53130}=0.38$

$P(k=1)=\dfrac{\dbinom{4}{1}\cdot \dbinom{25-4}{5-1}}{\dbinom{25}{5}}=\dfrac{\dbinom{4}{1}\cdot \dbinom{21}{4}}{\dbinom{25}{5}}\\\\\\P(k=1)=\dfrac{4\cdot 5985}{53130}=0.45$

$P(k=2)=\dfrac{\dbinom{4}{2}\cdot \dbinom{25-4}{5-2}}{\dbinom{25}{5}}=\dfrac{\dbinom{4}{2}\cdot \dbinom{21}{3}}{\dbinom{25}{5}}\\\\\\P(k=2)=\dfrac{6\cdot 1330}{53130}=0.15$