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3 years ago

Help! ANSWER ALL! Please!

1 answer:

8 0

So I am using Method 1
4 1/2^2ft
4*2+1=9/2

9/2÷13
9/2÷13/1
13/1÷2/9
cross out 13 & 9
=2/1

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Sam and Bethan share £54 in the ratio 5:4 how much dose each get

Agata [3.3K]

Ratio 5:4
Total shares = 5+4 = 9

then to find for 1 share = 54/9 = 6

So Sam share = 6 x 5 = 30
Bethan share = 6 x 4 = 24

Poof: 30+24 = 54

3 0

3 years ago

What is the length of AC? 3 ft 4 ft 9 ft 18 ft

iragen [17]

Answer: The length of AC is 18 ft.

Step-by-step explanation:

By the given diagram,

AM = MB and CN = NB

M and N are the mid points of the sides AB and CB respectively,

Thus, by the mid point theorem,

MN ║ AC,

By the alternative interior angle theorem,

∠BMN ≅ ∠BAC

∠BNM ≅ ∠BCA

Thus, by AA similarity postulate,

ΔBMN ≅ ΔBAC

By the property of similar triangles,

$\frac{BM}{BA}=\frac{MN}{AC}$

$\frac{BM}{BM+MA}=\frac{MN}{AC}$

$\frac{4}{4+4}=\frac{9}{AC}$

$\frac{4}{8}=\frac{9}{AC}$

$4AC=72\implies AC = 18\text{ ft}$

Thus, The length of AC is 18 ft.

5 0

3 years ago

Read 2 more answers

Of the 80 students in the student body, 60 voted for the winning candidate. What percent voted for the winning candidate

Olin [163]

Answer:

20 studenti fac laba on a romania

6 0

3 years ago

Read 2 more answers

Ninety-one percent of products come off the line within product specifications. Your quality control department selects 15 produ

Allisa [31]

Answer:

Probability of stopping the machine when $X < 9$ is 0.0002

Probability of stopping the machine when $X < 10$ is 0.0013

Probability of stopping the machine when $X < 11$ is 0.0082

Probability of stopping the machine when $X < 12$ is 0.0399

Step-by-step explanation:

There is a random binomial variable $X$ that represents the number of units come off the line within product specifications in a review of $n$ Bernoulli-type trials with probability of success $0.91$ . Therefore, the model is ${15 \choose x} (0.91) ^ {x} (0.09) ^ {(15-x)}$ . So:

$P (X < 9) = 1 - P (X \geq 9) = 1 - [{15 \choose 9} (0.91)^{9}(0.09)^{6}+...+{ 15 \choose 15}(0.91)^{15}(0.09)^{0}] = 0.0002$

$P (X < 10) = 1 - P (X \geq 10) = 1 - [{15 \choose 10}(0.91)^{10}(0.09)^{5}+...+{15 \choose 15} (0.91)^{15}(0.09)^{0}] = 0.0013$

$P (X < 11) = 1 - P (X \geq 11) = 1 - [{15 \choose 11}(0.91)^{11}(0.09)^{4}+...+{15 \choose 15} (0.91)^{15}(0.09)^{0}] = 0.0082$

$P (X < 12) = 1- P (X \geq 12) = 1 - [{15 \choose 12}(0.91)^{12}(0.09)^{3}+...+{15 \choose 15} (0.91)^{15}(0.09)^{0}] = 0.0399$

Probability of stopping the machine when $X < 9$ is 0.0002

Probability of stopping the machine when $X < 10$ is 0.0013

Probability of stopping the machine when $X < 11$ is 0.0082

Probability of stopping the machine when $X < 12$ is 0.0399

8 0

3 years ago

The rate at which a duck gains weight is proportional to the difference between its adult weight and its current weight. At time

kogti [31]

DB/dt = 1/5(100 - B)
5/(100 - B) dB = dt
-5 ln (100 - B) = t + C
Since B(0) = 20, then
-5 ln (100 - 20) = C
i.e. C = -5 ln 80

Thus -5 ln (100 - B) = t - 5 ln 80
or t = 5 ln 80 - 5 ln (100 - B) = 5 ln (80 / (100 - B))

When the weight = 40 grams
t = 5 In (80 / 100 - 40) = 5 In (80 / 60) = 5 ln (4/3) = 1.438 days
Rate of weight gain = 40 / 1.438 = 27.8 grams per day

When the weight = 70 grams
t = 5 ln (80 / 100 - 70) = 5 ln (80 / 30) = 5 ln (8/3) = 4.9 days
Rate of weight gain = 70/4.9 = 14.27 grams per day.

Therefore, the duck gains weight faster when it weighs 40 grams.

6 0

3 years ago