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3 years ago

Help with number 6 please!

Mathematics

1 answer:

hjlf3 years ago

6 0

Since the base price for membership with 0 classes is $30 and the price goes up by 5 dollars with each class, then the equation for y would be...

y = 30 + 5x

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assume that when adults with smartphones are randomly selected, 44% use them in meetings or classes. If 10 adult smartphones use

tensa zangetsu [6.8K]

Solution: The given random experiment follows Binomial distribution with $n=10,p=0.44$

Let $X$ be the number of adults who use their smartphones in meetings or classes.

Therefore, we have to find:

$P(X$

We know the binomial model is:

$P(X=x)=\binom{n}{x} p^{x} (1-p)^{n-x}$

$\therefore P(X$

$=\binom{10}{0}0.44^{0}(1-0.44)^{12-0}+\binom{10}{1}0.44^{1}(1-0.44)^{10-1}+\binom{10}{2}0.44^{2}(1-0.44)^{10-2}$

$=1 \times 1 \times 0.0030 + 10 \times 0.44 \times 0.0054 + 45 \times 0.1936 \times 0.009672$

$=0.0030+0.0238+0.0843$

$=0.1111$

Therefore, the probability that fewer than 3 of them is 0.1111

7 0

3 years ago

The sum of two segments are 75 cm, and one the segments are four times the other. Calculate the measurement of each segment

iVinArrow [24]

The length of one segment is 15 cm and other segment is 60 cm

Step-by-step explanation:

Let x be the length of one segment and y be the length of other segment

Then according to given statement

x+y=75 Eqn 1

And

one the segments are four times the other

x=4y Eqn 2

$Putting\ x=4y\ in\ equation\ 1$

$4y+y=75\\5y=75\\Dividing\ both\ sides\ by\ 5\\\frac{5y}{5}=\frac{75}{5}\\y=15\\Putting\ y=15\ in\ eqn\ 2\\x=4(15)\\x=60$

Hence,

The length of one segment is 15 cm and other segment is 60 cm

Keywords: Linear Equation, Variables

Learn more about linear equations at:

#LearnwithBrainly

8 0

3 years ago

FOR 15 POINTS Determine whether each given value of h makes the inequality 4h + 13 ≤ −7 true. Select Yes or No for each value.

Norma-Jean [14]

Answer:

Yes, no, yes

Step-by-step explanation:

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2 years ago

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AlexFokin [52]

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2 years ago

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How can I divide larger dividends.

Sphinxa [80]

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3 years ago

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