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hichkok12 [17]
2 years ago
10

. Mason had $24 to spend on 5 snacks. After buying them he had $4 left. How much did each

Mathematics
2 answers:
murzikaleks [220]2 years ago
6 0

Answer:

4

Step-by-step explanation:

5 times 4 is 20, which leaves 4

EleoNora [17]2 years ago
4 0

Answer:

$4

Step-by-step explanation:

Madison spent $20 because she had $4 left.

24-4=20

She used the $20 for 5 snacks

20 divided by 5 = 4

Each snack cost $4

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Solve the inequality.<br> |3x+7| &lt;_ 4
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Answer:

Using ∣x∣>a⇒x<−a or x>a, we get

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7 0
3 years ago
Let D be the event that a randomly chosen person has seen a dermatologist. Let S be the event that a randomly chosen person has
SSSSS [86.1K]

Given:

D be the event that a randomly chosen person has seen a dermatologist.

S be the event that a randomly chosen person has had surgery for skin cancer.

To find:

The correct notation for the probability that a randomly chosen person has had surgery for skin cancer, given that the person has seen a dermatologist.

Solution:

Conditional probability: Probability of A given B is:

P(A|B)=\dfrac{P(A\cap B)}{P(B)}

Let D be the event that a randomly chosen person has seen a dermatologist.

Let S be the event that a randomly chosen person has had surgery for skin cancer.

Using the conditional probability, the correct notation for the probability that a randomly chosen person has had surgery for skin cancer, given that the person has seen a dermatologist is P(S|D).

Therefore, the correct option is D.

8 0
2 years ago
ALGEBRA QUESTION PLS HELPS
cluponka [151]

The value of x is –7.

Solution:

Given expression:

$\left(\frac{1}{x+3}+\frac{6}{x^{2}+4 x+3}\right) \cdot \frac{x+3}{x+1}

Let us factor x^2+4x+3.

x^2+4x+3=(x+1)(x+3)

Substitute this in the fraction.

$\left(\frac{1}{x+3}+\frac{6}{(x+1)(x+3)}\right) \cdot \frac{x+3}{x+1}

To make the denominator same, multiply and divide the first term by (x +1).

$\left(\frac{(x+1)}{(x+1)(x+3)}+\frac{6}{(x+1)(x+3)}\right) \cdot \frac{x+3}{x+1}

Denominators are same, you can add the fractions.

$\left(\frac{x+1+6}{(x+1)(x+3)}\right) \cdot \frac{x+3}{x+1}

$\frac{x+7}{(x+1)(x+3)} \cdot \frac{x+3}{x+1}

Cancel the common term in the numerator and denominator.

$\frac{x+7}{x+1} \cdot \frac{1}{x+1}

Multiply the fractions.

$\frac{x+7}{(x+1)^2}

$\frac{x+7}{x^2+2x+1}

The expression is simplified to one rational expression.

Suppose the expression is equal to 0.

$\frac{x+7}{x^2+2x+1}=0

Do cross multiplication.

${x+7}=0\times (}{x^2+2x+1})

Any number or variable multiplied by 0 gives 0.

${x+7}=0

Subtract 7 from both sides of the equation.

${x+7-7}=0-7

x = –7

The value of x is –7.

7 0
3 years ago
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