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Lubov Fominskaja [6]
3 years ago
5

~Please help as soon as possible~

Mathematics
1 answer:
Sindrei [870]3 years ago
7 0
Question 1

well the is a 1:4  Faculty:trainee ratio
if there are   2:20, that would be a 1:10

you just need to find 1/4th of 20 and thats the total
20/4=5
You need five faculty total so
YOU NEED 3 MORE

Question 2

600 pretzels/10 minutes
?pretzels/52 minutes
600x52/10=3120

Question 3

700 people/ 2 tons
20,850,000/ ? tons
20,850,000 x 2 / 700= <span>59,571.4285714 OR about 60,000 tons</span>

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suppose you choose a marble from a bag containing 3 red marbles, 3 white marbles, and 5 blue marbles. You return the first marbl
Marta_Voda [28]

Answer:

15/121 is the answer.

8 0
3 years ago
4+2(7)= <br><br> 62+8×3= <br><br> 3(2+5)−5(3)+8= <br><br> 24−6+12÷2×3=
Temka [501]

4+2(7)= 18

62+8x3= 86

3(2+5)-5(3)+8= 14

24-6+12÷2x3= 36

Hope this helps you! :D

3 0
3 years ago
Help plsss
andrew11 [14]

(b) is the answer.

Step-by-step explanation:

By the Pythagorean Theorem,

A² + B² = C²

Where:

A = Length of side 1

B = Length of side 2

C = Hypotenuse

This rule applies to all right-angled triangles.

The length of the hypotenuse of a right-angled triangle is always the largest value.

Therefore, we can test the answers with the equation above.

(a)

8² + 18² = 20²

64 + 324 = 400

388 ≠ 400

The rule of Pythagorean theorem doesn't work on a, so (a) is not a right-angled triangle.

(b)

12² + 35² = 37²

144 + 1225 = 1369

1369 = 1369

The rule of Pythagorean theorem works here, so (b) is a right-angled triangle.

8 0
2 years ago
In a certain communications system, there is an average of 1 transmission error per 10 seconds. Assume that the distribution of
Nitella [24]

Answer:

The probability of 1 error in a period of 0ne - half minute is 0.1494

Step-by-step explanation:

Formula for poisson distribution:

P (X = k) = \frac{\exp^{- \lambda} \lambda^{x}  }{x!}

If there is an average of 1 error in 10 seconds

In one-half minutes (i.e. 30 seconds), there will be an average of 30/10 errors = 3 errors

\lambda = 3 errors\\x = 1

P (X = 1) = \frac{\exp^{-3} 3^{1}  }{1!}

1! = 1

P (X = 1) = 3 \exp^{-3}

P(X = 1) = 3 * 0.0498

P(X = 1) = 0.01494

7 0
3 years ago
Read 2 more answers
A botanist wishes to estimate the typical number of seeds for a certain fruit. She samples 48 specimens and counts the number of
baherus [9]

Answer: The open interval would be (31.4,42.5).

Step-by-step explanation:

Since we have given that

mean = 36.9

Standard deviation = 16.5

n = 48

At 98% confidence interval, z = 2.33

So, Interval would be

\bar{x}\pm z\dfrac{\sigma}{\sqrt{n}}\\\\=36.9\pm 2.33\dfrac{16.5}{\sqrt{48}}\\\\=36.9\pm 5.549\\\\=(36.9-5.5,36.9+5.6\\\\=(31.4,42.5)

Hence, the open interval would be (31.4,42.5).

5 0
3 years ago
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