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3 years ago

Hello can anyone help me with this ? Thank you ill mark brainless

Mathematics

2 answers:

Kruka [31]3 years ago

7 0

Answer:

$\huge\boxed{n=5m}$

$\huge\boxed{55 \ \text{novels}}$

Step-by-step explanation:

With this word problem, we can model it into an equation which we can interpret to find values.

If we need to read 5 novels per month and n represents the amount of novels read in m months, we know that n will be the dependent variable (y) as it's a direct function of the number of months. This makes m the independent variable (x).

Since there are 5 novels a month, the equation will be 5 times the number of months equals the number of novels per month.

aka - $n=5m$

We can now substitute inside the equation with 11 months as m, and find the value of n.

$n = 5 \cdot 11$
$n=55$

Hope this helped!

Alborosie3 years ago

3 0

Answer:

The answer is 55

Step-by-step explanation:

Because for every 1 month he reads 5 books

if you do 5 times 11 you get 55

and to check your answer you divide 55 by 5 you get 11

so your equation would be 5 times 11

and your answer would be 55.

hope this helps ;)

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Of all customers purchasing automatic garage-door openers, 75% purchase Swedish model. Let X = the number among the next 15 purc

Lelechka [254]

Answer:

$P(X=k) = {15 \choose k} * 0.75^{k}*0.25^{15-k}$

For any integer k between 0 and 15, and 0 for other values of k.

$P(X>10) = 0.2252+ 0.2252+ 0.1559+0.0668+0.0134 = 0.6865$

c) P(6 ≤ X ≤ 10) = 0.2737

d) μ = 15*0.75 = 11.25. σ² = 11.25*0.25 = 2.8125

Step-by-step explanation:

X is a binomial random variable with parameters n = 15, p = 0.75. Therefore

$P(X=k) = {15 \choose k} * 0.75^{k}*0.25^{15-k}$

For any integer k between 0 and 15, and 0 for other values of k.

P(X>10) = P(X=11) + P(X=12)+ P(X=13)+P(X=14)+P(x=15)

$P(X=11) = {15 \choose 11} * 0.75^{11} * 0.25^4 = 0.2252$

$P(X=12) = {15 \choose 12} * 0.75^{12} * 0.25^3 = 0.2252$

$P(X=13) = {15 \choose 13} * 0.75^{13} * 0.25^2 = 0.1559$

$P(X=14) = {15 \choose 14} * 0.75^{14} * 0.25 = 0.0668$

$P(X=15) = {15 \choose 15} * 0.75^{15} = 0.0134$

Thus,

$P(X>10) = 0.2252+ 0.2252+ 0.1559+0.0668+0.0134 = 0.6865$

c) P(6 ≤ X ≤ 10) = P(X = 6) + P(X = 7) + P(X = 8) + P(X=9) + P(X=10)

$P(X=6) = {15 \choose 6} * 0.75^{6} * 0.25^9 = 0.0034$

$P(X=7) = {15 \choose 7} * 0.75^{7} * 0.25^8 = 0.0131$

$P(X=8) = {15 \choose 8} * 0.75^{8} * 0.25^7 = 0.0393$

$P(X=9) = {15 \choose 9} * 0.75^{9} * 0.25^6 = 0.0918$

$P(X=10) = {15 \choose 10} * 0.75^{10} * 0.25^{5} = 0.1652$

Thereofre,

$P(6 \leq X \leq 10) = 0.0034 + 0.0134 + 0.0393 + 0.0918 + 0.1652 = 0.2737$

d) μ = n*p = 15*0.75 = 11.25

σ² = np(1-p) = 11.25*0.25 = 2.8125

3 0

3 years ago

38" points please help

nikdorinn [45]

1. The winner threw the ball 175 ft, so what we want to do now is we convert (175) ft into yd.

We already know that 1ft= 1/3 yd, so the distance in yd is:
175 ft* (1/3 yd/ 1 ft) = 58 1/3 yd.

The answer is A. 58 1/3 yards.

2. 75 min* (4.5 miles/ 15 min)= 22.5 miles.

The correct answer is D. 22.5

3. Let's say that there are 4 weeks in a month. There are 5 weekdays in a week, so the number of weekdays in a month is:
5 weekdays/ 1 week* (4 weeks/ 1 month)= 20 weekdays/ 1 month.

Alex runs 5.9 miles/ 1 weekday, so the number of miles he runs every month is:
(5.9 miles/ 1 weekday)* (20 weekdays/ 1 month)= 118 miles/ month.

The nearest number to 118 is 120, so the answer is D.120 miles

4. 10 sweaters to 8 shirts:

10:8= 10/8= 5/4

The answer is B. 5/4

Hope this would help~

3 0

3 years ago

Sharmila had 5/6 portion of the cake. She gave 2/6 portion of the whole cake to

zlopas [31]

Answer: $\dfrac{3}{6}$