Please pls plsss help me check 15 and 16

Ganezh [65]

Answer:

16

Step-by-step explanation:

Let the 1st part of your answer be x , so the 2nd part will be 40-x . From the given information, we can write the equation: (1/4)x = (3/8) × (40-x) . We can simplify this into (1/4)x = (120-3x)/8 ; 8x = 480-12x ; 8x+12x = 480 ; 20x = 480 ; x = 480/20; x = 24

Therefore, the 1st part = 24

Plug this into your 40-x equation to get: 40 - 24 = 16

3 0

3 years ago

Tim,Griff and Kay baked chocolate for Kay s birthday party.Tim baked 3 times as many as Kay.Griff baked half as many as Tim.Alto

lianna [129]

Answer: 48

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Define the number of cakes they baked in term of x
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Let the number of cakes Kay baked be x

Kay = x
Tim = 3x   ← Tim baked 3 times as many as Kay
Griff = 1.5x   ← Griff baked 1/2 as many as Tim

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Construct the equation and solve for x
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The total number of cakes baked was 88.

x + 3x + 1.5x = 88   ← Combine like terms
5.5 x = 88 ← Divide by 5.5
÷5.5   ÷5.5
x = 16

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Find the number of cakes each baked
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Kay = x = 16
TIm = 3x = 16 x 3 = 48
Griff = 1.5x = 1.5(16) = 24

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Answer: Tim baked 48 cakes.
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3 0

3 years ago

Simplify the expression 21z - 6z + 13

Bess [88]

Answer:

15z+13

Step-by-step explanation:

hey! all we have to do is combine like terms.

21z-6z+13

15z+13

there's your answer.

3 0

3 years ago

Read 2 more answers

A hospital’s financial reimbursement payor mix indicates 35% of the patients have Medicare and 15% have Medicaid. Of those who h

Damm [24]

Using conditional probability, it is found that there is a 0.035 = 3.5% probability that a hospital patient has both Medicare and Medicaid.

<h3>What is Conditional Probability?</h3>

<em>Conditional probability</em> is the <u>probability of one event happening, considering a previous event</u>. The formula 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 problem, the events are:

Event A: Patient has Medicare.

Event B: Patient has Medicaid.

For the probabilities, we have that:

35% of the patients have Medicare, hence $P(A) = 0.35$ .

Of those who have Medicare, there is a 10% chance they also have Medicaid, hence $P(B|A) = 0.1$ .

Then, applying the <em>conditional </em>probability:

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

$0.1 = \frac{P(A \cap B)}{0.35}$

$P(A \cap B) = 0.35(0.1)$

$P(A \cap B) = 0.035$

0.035 = 3.5% probability that a hospital patient has both Medicare and Medicaid.

You can learn more about conditional probability at brainly.com/question/14398287

4 0

2 years ago

Marking brainliest

sertanlavr [38]

Answer:

<h3>They fought is very major Americans battle war ll</h3><h3>Between 400,000 and 500,000 Hispanic Americans served in the US armed forces</h3><h3> during world War II out of total of</h3><h3>16,000,000 constituting 3.1% to 3.2% of the US Armed forces </h3><h2>I hope you like it </h2><h2>Please give me a brainliest answer </h2>

5 0

3 years ago

Please pls plsss help me check 15 and 16​

Please pls plsss help me check 15 and 16