Como fasso essa conta 2x-10=50

Mathematics

1 answer:

masya89 [10]3 years ago

8 0

Answer:

correct

Step-by-step explanation:

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Exit A five-sided solid has the numbers 1, 2, 3, 4, and 5. What is the probability of rolling two five-sided number solids and g

STALIN [3.7K]

To get a sum of 3, you must roll a 1 and a 2. To get a sum of 8, you must roll a 5 and 3, or two 4's.
There are two ways of rolling a 1 and a 2, two ways of rolling a 5 and a 3, and one way to roll two 4's.
There are $5^2=25$ possible outcomes when rolling two such solids.

So the probability of getting a sum of 3 or 8 is $\dfrac{2+2+1}{25}=\dfrac15$

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3 years ago

Subtract 2x^2-3xy+4y^2 from 3x^2-5xy-6y^2

Diano4ka-milaya [45]

Answer:

$-x^{2} +2xy+10y^{2}$

Step-by-step explanation:

$2x^2-3xy+4y^2-(3x^2-5xy-6y^2)\\2x^2-3xy+4y^2-3x^2+5xy+6y^2\\-x^{2} +2xy+10y^{2}$

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3 years ago

Help! What is |-3|+|3| ?

kvv77 [185]

$|a|= \left\{\begin{array}{ccc}a&for\ a\geq0\\-a&for\ a \ \textless \ 0\end{array}\right\\\\Examples:\\|2|=2\ because\ 2\geq0\\|0|=0\ because\ 0\geq0\\|-5|=-(-5)=5\ because\ -5 \ \textless \ 0\\|-100|=-100\ because\ -100 \ \textless \ 0$

|-3| + |3| = 3 + 3 = 6

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3 years ago

A researcher was interested in seeing how many names a class of 38 students could remember after playing a name game After playi

Andreas93 [3]

Answer:

Proportion of the students recalled more than 15 names is 91.77%.

Step-by-step explanation:

We are given that a researcher was interested in seeing how many names a class of 38 students could remember after playing a name game After playing the name game, the students were asked to recall as many first names of fellow students as possible.

The mean number of names recalled was 19.41 with a standard deviation of 3.17.

Let X = number of names recalled

SO, X ~ N( $\mu = 19.41,\sigma^{2} = 3.17^{2}$ )

The z-score probability distribution is given by ;

Z = $\frac{X-\mu}{\sigma} } }$ ~ N(0,1)

where, $\mu$ = mean number of names recalled = 19.41

$\sigma$ = standard deviation = 3.17

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, proportion of the students recalled more than 15 names is given by = P(X > 15 names)

P(X > 15) = P( $\frac{X-\mu}{{\sigma} } }$ > $\frac{15-19.41}{3.17} }$ ) = P(Z > -1.39)