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3 years ago

6

PLS HELP AGAIN ASAP THIS THE SECOND TO LAST PROBLEM ON MORE AFTER THIS

1 answer:

Rufina [12.5K]3 years ago

3 0

Answer:

140

Step-by-step explanation:

because 2 cups of flour turned to 20 cups is to 14 cups of sugar to 140 cups

so the ratio goes from 2: 14 to 20: 140

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Suppose that a department contains 9 men and 15 women. How many different committees of 6 members are possible if the committee

Drupady [299]

Answer: The required number of different possible committees is 81172.

Step-by-step explanation: Given that a department contains 9 men and 15 women.

We are to find the number of different committees of 6 members that are possible if the committee must have strictly more women than men.

Since we need committees of 6 members, so the possible combinations are

(4 women, 2 men), (5 women, 1 men) and (6 women).

Therefore, the number of different committees of 6 members is given by

$n\\\\\\=^{15}C_4\times ^9C_2+^{15}C_5\times ^9C_1+^{15}C_6\\\\\\=\dfrac{15!}{4!(15-4)!}\times \dfrac{9!}{2!(9-2)!}+\dfrac{15!}{5!(15-5)!}\times \dfrac{9!}{1!(9-1)!}+\dfrac{15!}{6!(15-6)!}\\\\\\\\=\dfrac{15\times14\times13\times12\times11!}{4\times3\times2\times1\times11!}\times\dfrac{9\times8\times7!}{2\times1\times7!}+\dfrac{15\times14\times13\times12\times11\times10!}{5\times4\times3\times2\times1\times10!}\times\dfrac{9\times8!}{1\times8!}+\dfrac{15\times14\times13\times12\times11\times10\times9!}{6\times5\times4\times3\times2\times1\times9!}\\\\\\=1365\times36+3003\times9+5005\\\\=49140+27027+5005\\\\=81172.$

Thus, the required number of different possible committees of 6 members is 81172.

8 0

3 years ago

A pack of candy bars has 6 bars. Each bar weighs 3.86 ounces. How much does the pack weigh?

san4es73 [151]

Answer:

23.16 punces

Step-by-step explanation:

6x3.86=23.16

3 0

3 years ago

Read 2 more answers

For the pair of vectors, find U.V.<br> U=-2i<br> V=4i

maks197457 [2]

Answer:

8

Step-by-step explanation:

If you want the dot product, here it is: (0)(0) + (-2i)(4i) = 8

6 0

4 years ago

What is the area of ABC?

notsponge [240]

Let D be the Intersection of Height of the Triangle and Base of the Triangle BC

From the Figure, We can notice that Triangle ADB is a Right angled Triangle.

We know that, In a Right angled Triangle :

$\bigstar$ (Hypotenuse)² = (First Leg)² + (Second leg)²

In Triangle, ADB : AB is the Hypotenuse, AD is the First leg and BD is the Second leg

Given : AB = 15 and BD = 9

Substituting the values, We get :

$:\implies$ (15)² = (AD)² + (9)²

$:\implies$ 225 = (AD)² + 81

$:\implies$ (AD)² = 225 - 81

$:\implies$ (AD)² = 144

$:\implies$ (AD)² = (12)²

$:\implies$ AD = 12

We know that, In a Right angled Triangle :

$\bigstar\;\;\boxed{\mathsf{Tan\theta = \dfrac{Opposite\;Side}{Adjacent\;Side}}}$

Now, Consider Triangle ADC : With respect to 45° Angle, AD is the Opposite Side and DC is the Adjacent Side

$:\implies \mathsf{In\;Triangle\;ADC,\;Tan45^{\circ} = \dfrac{AD}{DC}}$

$\mathsf{:\implies 1 = \dfrac{12}{DC}}$

$:\implies$ DC = 12

$:\implies$ Total Length of the Base (BC) = BD + DC

$:\implies$ Total Length of the Base (BC) = 9 + 12

$:\implies$ Total Length of the Base (BC) = 21

We know that, Area of the Triangle is given by :

$\bigstar\;\;\boxed{\mathsf{Area = \dfrac{1}{2} \times Base \times Height}}$

In Triangle, ABC : AD is the Height and BC is the Base

$:\implies \mathsf{Area\;of\;the\;Triangle\;ABC = \dfrac{1}{2} \times BC \times AD}$

$:\implies \mathsf{Area\;of\;the\;Triangle\;ABC = \dfrac{1}{2} \times 21 \times 12}$

$:\implies \mathsf{Area\;of\;the\;Triangle\;ABC = (21 \times 6)}$

$:\implies \mathsf{Area\;of\;the\;Triangle\;ABC = 126}$

7 0

4 years ago

3 1/7 x 10 answer plz

Yanka [14]

Answer:

31.43

Step-by-step explanation:

3 1/7 = 3.14285714285

3 1/7 x 10 = 31.4285714286 which is about 31.43

4 0

3 years ago

Read 2 more answers