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1 year ago

6

If a certain number of Grade IV students share 28 or 39 skill books in Mathematics,there are remainders of 4 and 3 books. What i

s the largest possible number of students?

1 answer:

nadya68 [22]1 year ago

7 0

Answer:

12 students

Step-by-step explanation:

<u>Below numbers are divisible by the number of students:</u>

28 - 4 = 24
39 - 3 = 36

<u>Lets find GCF of 24 and 36:</u>

24 = 2*2*2*3
36 = 2*2*3*3
GCF(24,36) = 2*2*3 = 12

The largest possible number of students is 12

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Eli needs pieces of ribbon cut into strips that are 2 1\2 feet long. The roll of ribbon is 7 1\2 feet long. Which expression cou

julsineya [31]

Answer: There are 3 strips that can be cut from the roll of ribbon.

Step-by-step explanation:

since we have given that

Length of a ribbon is given by

$7\frac{1}{2}\ ft\\\\=\frac{15}{2}\ ft$

Length of pieces of ribbon cut into strips is given by

$2\frac{1}{2}\ ft\\\\=\frac{5}{2}\ ft$

So, we need to find the number of strips that can be cut is given by

$\text{ Number of strips }=\frac{\text{Length of roll}}{\text{ Length of strip}}\\\\=\frac{\frac{15}{2}}{\frac{5}{2}}\\\\=\frac{15\times 2}{2\times 5}\\\\=\frac{15}{5}\\\\=3$

Hence, there are 3 strips that can be cut from the roll of ribbon.

6 0

2 years ago

Read 2 more answers

The number of flaws in a fiber optic cable follows a Poisson distribution. It is known that the mean number of flaws in 50m of c

boyakko [2]

Answer:

(a) The probability of exactly three flaws in 150 m of cable is 0.21246

(b) The probability of at least two flaws in 100m of cable is 0.69155

(c) The probability of exactly one flaw in the first 50 m of cable, and exactly one flaw in the second 50 m of cable is 0.13063

Step-by-step explanation:

A random variable X has a Poisson distribution and it is referred to as Poisson random variable if and only if its probability distribution is given by

$p(x;\lambda)=\frac{\lambda e^{-\lambda}}{x!}$ for x = 0, 1, 2, ...

where $\lambda$ , the mean number of successes.

(a) To find the probability of exactly three flaws in 150 m of cable, we first need to find the mean number of flaws in 150 m, we know that the mean number of flaws in 50 m of cable is 1.2, so the mean number of flaws in 150 m of cable is $1.2 \cdot 3 =3.6$

The probability of exactly three flaws in 150 m of cable is

$P(X=3)=p(3;3.6)=\frac{3.6^3e^{-3.6}}{3!} \approx 0.21246$

(b) The probability of at least two flaws in 100m of cable is,

we know that the mean number of flaws in 50 m of cable is 1.2, so the mean number of flaws in 100 m of cable is $1.2 \cdot 2 =2.4$

$P(X\geq 2)=1-P(X$

$P(X\geq 2)=1-p(0;2.4)-p(1;2.4)\\\\P(X\geq 2)=1-\frac{2.4^0e^{-2.4}}{0!}-\frac{2.4^1e^{-2.4}}{1!}\\\\P(X\geq 2)\approx 0.69155$

(c) The probability of exactly one flaw in the first 50 m of cable, and exactly one flaw in the second 50 m of cable is

$P(X=1)=p(1;1.2)=\frac{1.2^1e^{-1.2}}{1!}\\P(X=1)\approx 0.36143$

The occurrence of flaws in the first and second 50 m of cable are independent events. Therefore the probability of exactly one flaw in the first 50 m and exactly one flaw in the second 50 m is

$(0.36143)(0.36143) = 0.13063$

4 0

2 years ago

James folds a piece of paper in half several times,each time unfolding the paper to count how many equal parts he sees. After fo

Snezhnost [94]

Answer:

There will be total 2048 parts of the given paper if James if able to fold the paper eleven times.

The needed function is $y = 2 ^n$

Step-by-step explanation:

Let us assume the piece of paper James decides to fold is a SQUARE.

Now, let us assume:

n : the number of times the paper is folded.

y : The number of parts obtained after folds.

Now, if the paper if folded ONCE ⇒ n = 1

Also, when the pap er is folded once, the parts obtained are TWO equal parts.

⇒ for n = 1 , y = 2 ..... (1)

Similarly, if the paper if folded TWICE ⇒ n = 2

Also, when the paper is folded twice, the parts obtained are FOUR equal parts.

⇒ for n = 2 , y = 4 ..... (2)

⇒ $y = 2^2 = 2^n$

Continuing the same way, if the paper is folded SEVEN times ⇒ n = 7

So, $y = 2^ n = 2^7 = 128$

⇒ There are total 128 equal parts.

Lastly, if the paper is folded ELEVEN times ⇒ n = 11

So, $y = 2^ n = 2^{11} = 2048$

⇒ There are total 2048 equal parts.

Hence, there will be total 2048 parts of the given paper if James if able to fold the paper eleven times.

And the needed function is $y = 2 ^n$

8 0

2 years ago

Solve y=3bx-7x for x

irina [24]

Y = 3bx - 7x
y = x(3b - 7)

Divide each side by 3b - 7 (assume that it is not zero).
$\frac{y}{3b-7} =x$

Answer:
$x= \frac{y}{3b-7}$

6 0

2 years ago

A bridge is rated to a capacity of 100 British tons. What is the maximum weight the bridge can support in kilograms? (Round to t

S_A_V [24]

Maximum weight the bridge can support in kilograms is 101696

Step-by-step explanation:

Step 1: Given capacity of bridge = 100 British tons. Find how many kilograms are equivalent to 1 British ton.

1 British ton = 2240 pounds

1 pound = 0.454 kg

⇒ 1 British ton = 2240 × 0.454 kg = 1016.96 kg

Step 2: Find how many kilograms are in 100 British tons.

⇒ 100 × 1016.96 = 101696

3 0

2 years ago