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3 years ago

11

(AAAASSSSSSAAAAPPPP!)

2 answers:

Ierofanga [76]3 years ago

5 0

C. 3+6x because your contributing the equation

Paul [167]3 years ago

3 0

The answer to this question is C.

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Mary and Kyle together sold 225 candy bars for a fund raiser. Kyle sold twice as many candy bars as Mary.

NARA [144]

Answer:

Mary has 76.5, and Kyle has 178.5.

5 0

2 years ago

Read 2 more answers

at a particular high school, 42% of the students participate in sports and 25% of the students participate in drama. if 53% of t

baherus [9]

The probability that a student participates in both sports and drama is $0.14$ .

<h3>What is the formula for P(AUB), where A and B are any two events?</h3>

If $A$ and $B$ are any two events, then the probability of the joint event $(A\cup B)$ is given by the following formula: $P(A\cup B)=P(A)+P(B)-P(A\cap B)$

Given that 42% of the students participate in sports and 25% of the students participate in drama and 53% of the students participate in either sports or drama.

Suppose $S$ denotes that "a student participates in sports" and $D$ denotes that "a student participates in drama".
So, we have $P(S)=\frac{42}{100}=0.42$ , $P(D)=\frac{25}{100}=0.25$ , $P(S\cup D)=\frac{53}{100}=0.53$ .

We want to find the probability that a student participates in both sports and drama i.e., we want to find $P(S\cap D)$ .

By the above formula, we obtain:

$P(S\cup D)=P(S)+P(D)-P(S\cap D)\\\Longrightarrow P(S\cap D)=P(S)+P(D)-P(S\cup D)\\\Longrightarrow P(S\cap D)=0.42+0.25-0.53\\\therefore P(S\cap D)=0.14=\frac{14}{100}$

Therefore, the probability that a student participates in both sports and drama is $0.14$ .

To learn more about probability, refer: brainly.com/question/24756209

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6 0

2 years ago

Read 2 more answers

Find an equation for the graph

Elanso [62]

Answer:

y=6^x+0.612 - 4

Step-by-step explanation:

x = -4 ---> horizontal asymptote

m = 6 ---> use points (0, -1) and (1, 5)

Parent function of the graph: $y=b^x$

Our equation: $y=6^x$

Add alterations:

Reflections = N/A
Vertical & horizontal shifts = down 4
Vertical & horizontal stretches = left approx 0.612

Final equation: y=6^x+0.612 - 4

6 0

2 years ago

Help anyone can help me do this question,I will mark brainlest.

Ad libitum [116K]

Answer:

David walks 132m.

Using theorem of pythagoras and the dimensions of the kite to calculate. Refer to the picture.

Hope it helps.

3 0

3 years ago

Help math homework both questions

-Dominant- [34]

Answer:

Q-12 The vertical length of the top part of elevator to ground is 0.8484 meters .

Q -14 The glide runway path is 24,752.47 feet .

Step-by-step explanation:

Q - 12

Given as :

The diagonal length of the elevator = e = 1.2 meters

The vertical length of the top part of elevator to ground = x meters

The angle made by elevator with ground = Ф = 45°

<u>Now, According to figure</u>

Sin angle = $\dfrac{\textrm perpendicular}{\textrm hypotenuse}$

Or, Sin Ф = $\dfrac{\textrm x}{\textrm e}$

Or, Sin 45° = $\dfrac{\textrm x meters}{\textrm 1.2 meters}$

Or, 0.707 × 1.2 meters = x

∴ x = 0.8484 meters

So, The vertical length of the top part of elevator to ground = x = 0.8484 meters

Hence,The vertical length of the top part of elevator to ground is 0.8484 meters . Answer

Q-14

Given as:

The altitude of decent of plane = H = 10,000 feet

The angle of descend = Ф = 22°

Let the glide runway path = y feet

<u>Now, According to question</u>

Tan angle = $\dfrac{\textrm perpendicular}{\textrm base}$

Or, Tan Ф = $\dfrac{\textrm H}{\textrm y}$

Or, Tan 22° = $\dfrac{\textrm 10,000 feet}{\textrm y feet}$

Or, 0.4040 = $\dfrac{\textrm 10,000 feet}{\textrm y feet}$

Or, y = $\dfrac{\textrm 10,000 feet}{\textrm 0.4040}$

∴ y = 24,752.47

So,The glide runway path = y = 24,752.47 feet

Hence, The glide runway path is 24,752.47 feet . Answer

6 0

3 years ago