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2 years ago

My question is in the picture pls help

Mathematics

2 answers:

Viktor [21]2 years ago

8 0

Answer:

probally 80 or 90

Step-by-step explanation:

Anna007 [38]2 years ago

3 0

Answer:

109°

Step-by-step explanation:

subtract 180 from 71 and you will get 109°

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Ok this is getting stressful and hard already, we’re not even learning it yet but whatever, please help me will mark Brainliest

s2008m [1.1K]

Answer:

yes? im pretty sure yeah

Step-by-step explanation:

6 0

3 years ago

Use the diagram shown.

PSYCHO15rus [73]

Answer:

Step-by-step explanation:

8 0

3 years ago

Naval intelligence reports that 4 enemy vessels in a fleet of 17 are carrying nuclear weapons. If 9 vessels are randomly targete

icang [17]

Answer:

0.7588 = 75.88% probability that more than 1 vessel transporting nuclear weapons was destroyed

Step-by-step explanation:

The vessels are destroyed without replacement, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

The probability of x successes is given by the following formula:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

In which:

x is the number of successes.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

In this question:

Fleet of 17 means that $N = 17$

4 are carrying nucleas weapons, which means that $k = 4$

9 are destroyed, which means that $n = 9$

What is the probability that more than 1 vessel transporting nuclear weapons was destroyed?

This is:

$P(X > 1) = 1 - P(X \leq 1)$

In which

$P(X \leq 1) = P(X = 0) + P(X = 1)$

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = 0) = h(0,17,9,4) = \frac{C_{4,0}*C_{13,9}}{C_{17,9}} = 0.0294$

$P(X = 1) = h(1,17,9,4) = \frac{C_{4,1}*C_{13,8}}{C_{17,9}} = 0.2118$

Then

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.0294 + 0.2118 = 0.2412$

$P(X > 1) = 1 - P(X \leq 1) = 1 - 0.2412 = 0.7588$

0.7588 = 75.88% probability that more than 1 vessel transporting nuclear weapons was destroyed

8 0

2 years ago

Bingo uses the numbers 1-75. If you randomly select a number, what is the probability it will not be a multiple of ten? Explain

ValentinkaMS [17]

Multiples of ten are: 10, 20, 30, 40, 50, 60, & 70.

So there are (7 options that ARE multiplies of ten) out of (75 total options)

Thus, there are (75 - 7 options that are NOT multiples of ten) out of (75 total options).

Answer: $\frac{68}{75}$

7 0

3 years ago

The first answer already given, what's the second answer (in the green box)

chubhunter [2.5K]

Solution:

Since the graph passes through the given points, (7, 20) & (-2, 11) are the solutions of the given equation y = x + ?.

⇒(x, y) = (7, 20); (-2, 11)

Substituting the variables with (7, 20),

20 = 7 + ?

20 - 7 = ?

? = 13

Similarly,

11 = -2 + ?

? = 13

∴ y = 13

8 0

2 years ago