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ValentinkaMS [17]

2 years ago

5

Do all 4 pls it is easy

1 answer:

Novosadov [1.4K]2 years ago

5 0

Answer:

1. -2y

2.13x

3. 8d-3c

4. 8r+(-6s)

Step-by-step explanation:

You might be interested in

Answer the question belowwwww

Nutka1998 [239]

Answer:

They swam 15 days

Step-by-step explanation:

Divide 48 by 3.2

6 0

2 years ago

Brandon bought 21 chicken wings for $39.90. How much would it cost for 15 wings?

Pachacha [2.7K]

Answer:

$28.50

Step-by-step explanation:

21 chicken wings= $39.90

1 chicken wing= 39.90÷21= $1.90

15 chicken wings= 15×1.90= $28.50

4 0

3 years ago

A certain virus infects one in every 200 people. a test used to detect the virus in a person is positive 70% of the time when t

V125BC [204]

We're told that

$P(A)=\dfrac1{200}=0.005\implies P(A^C)=0.995$

$P(B\mid A)=0.7$

$P(B\mid A^C)=0.05$

a. We want to find $P(A\mid B)$ . By definition of conditional probability,

$P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}$

By the law of total probability,

$P(B)=P(B\cap A)+P(B\cap A^C)=P(B\mid A)P(A)+P(B\mid A^C)P(A^C)$

Then

$P(A\mid B)=\dfrac{P(B\mid A)P(A)}{P(B\mid A)P(A)+P(B\mid A^C)P(A^C)}\approx0.0657$

(the first equality is Bayes' theorem)

b. We want to find $P(A^C\mid B^C)$ .

$P(A^C\mid B^C)=\dfrac{P(A^C\cap B^C)}{P(B^C)}=\dfrac{P(B^C\mid A^C)P(A^C)}{1-P(B)}\approx0.9984$

since $P(B^C\mid A^C)=1-P(B\mid A^C)$ .

4 0

2 years ago

What is the product of all constants $k$ such that the quadratic $x^2 + kx +15$ can be factored in the form $(x+a)(x+b)$, where

Setler79 [48]

Answer:

$k=-16,k=-8,k=8,k=16$

Step-by-step explanation:

We are given quadratic equations as

$x^2+kx+15$

and it can be factored as

$=(x+a)(x+b)$

now, we can multiply factor term

$(x+a)(x+b)=x^2+(a+b)x+ab$

now, we can compare

$x^2+(a+b)x+ab=x^2+kx+15$

so, we get

$k=a+b$

$ab=15$

we are given that

'a' and 'b' are integers

so, we can find all possible factors

$15=(-1\times -15),(1\times 15)$

$15=(-3\times -5),(3\times 5)$

so, we can find k

At $(-1\times -15)$ :

$k=a+b$

we can plug values

$k=-1-15$

$k=-16$

At $(1\times 15)$ :

$k=a+b$

we can plug values

$k=1+15$

$k=16$

At $(-3\times -5)$ :

$k=a+b$

we can plug values

$k=-3-5$

$k=-8$

At $(3\times 5)$ :

$k=a+b$

we can plug values

$k=3+5$

$k=8$

So, values of k are

$k=-16,k=-8,k=8,k=16$

6 0

3 years ago

Identify the triangle that contains an acute angle for which the sine and cosine ratios are equal.

algol13

Answer:

i need the diagram

Step-by-step explanation:

plz

3 0

2 years ago

Read 2 more answers