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2 years ago

What is y=2x+1 3x+4y=26 equal?

Mathematics

1 answer:

strojnjashka [21]2 years ago

7 0

Substitute in y= equation into 2nd equation to start.
(2,5)

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Write the word or phrase that best completes each statement or answers the question.

aniked [119]

I believe that the answer is A

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2 years ago

A quadrilateral has two consecutive angles that measure 90 degrees each

hoa [83]

The answer is C)i, ii, iii, iv

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3 years ago

Read 2 more answers

Burka [1]

Answer:

Step-by-step explanation:

$F \alpha \frac{1}{d^{2} }$

$F = \frac{K}{d^{2} }$

When F = 18; d = 2

$18 = \frac{K}{2^{2} }$

$18 = \frac{K}{4}$

Cross multiply;

18 x 4 = K

72 = K

There the equation connecting F and $d^{2}$ is

$F = \frac{72}{d}$

Now, Find F when d = 6

All you do is to substitute d = 6 in to $F = \frac{72}{d}$

$F = \frac{72}{6}$

Therefore;

F = 12

Please mark me brainiest if correct.

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2 years ago

Benjamin drove a distance of 301.5 miles in

slava [35]

Answer: 67 miles per hour

Step-by-step explanation: $v = \frac{s}{t}$ where v is velocity, s is space and t is time;

$\frac{301.5 miles}{4.5 h}$ = 67

6 0

3 years ago

A diagnostic test for a disease is such that it (correctly) detects the disease in 90% of the individuals who actually have the

Ne4ueva [31]

Answer:

0.2177 = 21.77% conditional probability that she does, in fact, have the disease

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Test positive

Event B: Has the disease

Probability of a positive test:

90% of 3%(has the disease).

1 - 0.9 = 0.1 = 10% of 97%(does not have the disease). So

$P(A) = 0.90*0.03 + 0.1*0.97 = 0.124$

Intersection of A and B:

Positive test and has the disease, so 90% of 3%

$P(A \cap B) = 0.9*0.03 = 0.027$

What is the conditional probability that she does, in fact, have the disease

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.027}{0.124} = 0.2177$

0.2177 = 21.77% conditional probability that she does, in fact, have the disease

3 0

3 years ago