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liraira [26]
3 years ago
8

A craft weighs 6 kg when empty. A lemon weighs about 0.2 kg. For economical shipping the crate with lemons must weigh at least 4

5 kg. How many lemons should be put in the crate?
Mathematics
1 answer:
Anna71 [15]3 years ago
4 0
You would need 225 lemons in the crate. you just do 45 ÷ 0.2 to find the answer xx
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If a is less than 10 and b is less than 5,it must be true that
Misha Larkins [42]
D. The greatest a can be is 9 and the greatest b can be is 4. a+b= 13 which is less than 20
4 0
3 years ago
What is 413 divided by 7 please show explanation or picture if applicable.
professor190 [17]

Answer: 59 is your answer.

Step-by-step explanation:

Divide and multiply.

Subtract and compare. Be sure the number is less than the divisor.

Bring down the next number. Divide and multiply.

Subtract and compare. Be sure the number is less than the divisor.

3 0
3 years ago
Read 2 more answers
a packe of cards conatins 4 red and 5 black cards. a hand of 5 cards is drawn without replacment. What is the proablity of all f
Vladimir79 [104]

Answer: Our required probability is 0.3387.

Step-by-step explanation:

Since we have given that

Number of red cards = 4

Number of black cards = 5

Number of cards drawn = 5

We need to find the probability of getting exactly three black cards.

Probability of getting a black card = \dfrac{5}{9}

Probability of getting a red card = \dfrac{4}{9}

So, using "Binomial distribution", let X be the number of black cards:

P(X=3)=^5C_3(\dfrac{5}{9})^3(\dfrac{4}{9})^2\\\\P(X=3)=0.3387

Hence, our required probability is 0.3387.

5 0
3 years ago
Evaluate the double integral.
Fynjy0 [20]

Answer:

\iint_D 8y^2 \ dA = \dfrac{88}{3}

Step-by-step explanation:

The equation of the line through the point (x_o,y_o) & (x_1,y_1) can be represented by:

y-y_o = m(x - x_o)

Making m the subject;

m = \dfrac{y_1 - y_0}{x_1-x_0}

∴

we need to carry out the equation of the line through (0,1) and (1,2)

i.e

y - 1 = m(x - 0)

y - 1 = mx

where;

m= \dfrac{2-1}{1-0}

m = 1

Thus;

y - 1 = (1)x

y - 1 = x ---- (1)

The equation of the line through (1,2) & (4,1) is:

y -2 = m (x - 1)

where;

m = \dfrac{1-2}{4-1}

m = \dfrac{-1}{3}

∴

y-2 = -\dfrac{1}{3}(x-1)

-3(y-2) = x - 1

-3y + 6 = x - 1

x = -3y + 7

Thus: for equation of two lines

x = y - 1

x = -3y + 7

i.e.

y - 1 = -3y + 7

y + 3y = 1 + 7

4y = 8

y = 2

Now, y ranges from 1 → 2 & x ranges from y - 1 to -3y + 7

∴

\iint_D 8y^2 \ dA = \int^2_1 \int ^{-3y+7}_{y-1} \ 8y^2 \ dxdy

\iint_D 8y^2 \ dA =8 \int^2_1 \int ^{-3y+7}_{y-1} \ y^2 \ dxdy

\iint_D 8y^2 \ dA =8 \int^2_1  \bigg ( \int^{-3y+7}_{y-1} \ dx \bigg)   dy

\iint_D 8y^2 \ dA =8 \int^2_1  \bigg ( [xy^2]^{-3y+7}_{y-1} \bigg ) \ dy

\iint_D 8y^2 \ dA =8 \int^2_1  \bigg ( [y^2(-3y+7-y+1)]\bigg ) \ dy

\iint_D 8y^2 \ dA =8 \int^2_1  \bigg ([y^2(-4y+8)] \bigg ) \ dy

\iint_D 8y^2 \ dA =8 \int^2_1  \bigg ( -4y^3+8y^2 \bigg ) \ dy

\iint_D 8y^2 \ dA =8 \bigg [\dfrac{ -4y^4}{4}+\dfrac{8y^3}{3} \bigg ]^2_1

\iint_D 8y^2 \ dA =8 \bigg [ -y^4+\dfrac{8y^3}{3} \bigg ]^2_1

\iint_D 8y^2 \ dA =8 \bigg [ -2^4+\dfrac{8(2)^3}{3} + 1^4- \dfrac{8\times (1)^3}{3}\bigg]

\iint_D 8y^2 \ dA =8 \bigg [ -16+\dfrac{64}{3} + 1- \dfrac{8}{3}\bigg]

\iint_D 8y^2 \ dA =8 \bigg [ -15+ \dfrac{64-8}{3}\bigg]

\iint_D 8y^2 \ dA =8 \bigg [ -15+ \dfrac{56}{3}\bigg]

\iint_D 8y^2 \ dA =8 \bigg [  \dfrac{-45+56}{3}\bigg]

\iint_D 8y^2 \ dA =8 \bigg [  \dfrac{11}{3}\bigg]

\iint_D 8y^2 \ dA = \dfrac{88}{3}

4 0
2 years ago
3) Jacob has 4 granola bars. He cuts them into pieces that are 1/3 of a bar each. After he cuts the 4 bars, how many pieces does
klio [65]

Answer:

12

Step-by-step explanation:

if each bar is cut into 3 pieces, then we multiply 4 by 3 to get 12.

5 0
3 years ago
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