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3 years ago

Evaluate the algebraic expression 6g - 11 =8

Mathematics

1 answer:

Alexandra [31]3 years ago

7 0

The answer is to G is -2

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In a recent year, grade 8 Washington State public school students taking a mathematics assessment test had a mean score of 281 w

sweet-ann [11.9K]

Answer:

The probability that the mean test score is greater than 290

P(X⁻ > 290 ) = 0.0217

Step-by-step explanation:

Step(i):-

Mean of the Population (μ) = 281

Standard deviation of the Population = 34.4

Let 'X' be a random variable in Normal distribution

Given X = 290

$Z = \frac{x -mean}{\frac{S.D}{\sqrt{n} } } = \frac{290-281}{4.44} = 2.027$

Step(ii):-

The probability that the mean test score is greater than 290

P(X⁻ > 290 ) = P( Z > 2.027)

= 0.5 - A ( 2.027)

= 0.5 - 0.4783

= 0.0217

The probability that the mean test score is greater than 290

P(X⁻ > 290 ) = 0.0217

6 0

3 years ago

Two percent of Jennie’s skin cells were burned when she escaped from a fire. If 3.7×10^10 of her skin cells were burned then, ho

gulaghasi [49]

$1813\times10^9$ cells were not burnt when Jennie escaped from the fire.

Solution:

Given:

2% of skin cells were burnt i.e, $3.7\times10^{10}$ of skin cells were burnt.

To find:

Number of skin cells that were not burnt

We can write $3.7\times10^{10} \text{ as } 37\times10^{9}$

Therefore, by equating the values we get,

2% of the cell is $37\times10^9$

1% of the cell is $\frac{37\times10^9}{3}=18.5\times10^9$

As, 1% of the cell is $18.5\times10^9$

98% of the cell will be $98\times18.5\times10^9=1813\times10^9$

$\therefore \text{ skin cells that are not burnt is } 1813\times10^9$

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3 years ago