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2 years ago

This is not an equation but if I get a 100/100 on a lab and I have a C in the same class what will my grade be?

Mathematics

1 answer:

alukav5142 [94]2 years ago

5 0

Answer:

probably a B.

Step-by-step explanation:

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What is the greatest one between 95% , 9/10, 0.89

Zarrin [17]

Answer:

95%

Step-by-step explanation:

9/10 is the same as 90%; which is less than 95%

0.89 is the same as 89%; which is less than both 90% and 95%

Hope this helps! Please mark brainliest! Have a great day!

Thank you for using Brainly!

8 0

2 years ago

Read 2 more answers

Tourists from a tour bus were asked about places they visited during their stay in a city. The results are shown in the table.

Eddi Din [679]

The answer is probaly 60 % i think

3 0

3 years ago

1. What is the spinning mo tion of the earth called? = rotation

Tasya [4]

Answer:

<u><em>Rotation and Revolution are two motions of the earth.</em></u>

Step-by-step explanation:

The description is in the image below:

5 0

2 years ago

Solve the following equation using the quadratic formula.

nataly862011 [7]

Answer:

x=3.5,x=-6

I hope this help

7 0

2 years ago

An Airliner has a capacity for 300 passengers. If the company overbook a flight with 320 passengers, What is the probability tha

TEA [102]

Answer:

The probability is $P(X >300 ) = 0.97219$

Step-by-step explanation:

From the question we are told that

The capacity of an Airliner is k = 300 passengers

The sample size n = 320 passengers

The probability the a randomly selected passenger shows up on to the airport

$p = 0.96$

Generally the mean is mathematically represented as

$\mu = n* p$

=> $\mu = 320 * 0.96$

=> $\mu = 307.2$

Generally the standard deviation is

$\sigma = \sqrt{n * p * (1 -p ) }$

=> $\sigma = \sqrt{320 * 0.96 * (1 -0.96 ) }$

=> $\sigma =3.50$

Applying Normal approximation of binomial distribution

Generally the probability that there will not be enough seats to accommodate all passengers is mathematically represented as

$P(X > k ) = P( \frac{ X -\mu }{\sigma } > \frac{k - \mu}{\sigma } )$

Here $\frac{ X -\mu }{\sigma } =Z (The \ standardized \ value \ of \ X )$

=> $P(X >300 ) = P(Z > \frac{300 - 307.2}{3.50} )$

Now applying continuity correction we have

$P(X >300 ) = P(Z > \frac{[300+0.5] - 307.2}{3.50} )$

=> $P(X >300 ) = P(Z > \frac{[300.5] - 307.2}{3.50} )$

=> $P(X >300 ) = P(Z > -1.914 )$

From the z-table

$P(Z > -1.914 ) = 0.97219$

$P(X >300 ) = 0.97219$

8 0

2 years ago