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2 years ago

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HELP ASAP PLEASE!!!! 1. Create a unique equation using the graphic above with the solution (answer) of 36. There are many possib

le equations you could create to equal 36 using the chart above. 2. Describe the process you used to create your equation in prompt 1.

1 answer:

gtnhenbr [62]2 years ago

5 0

Answer:

just using normal x and y coordinates

4x + 2y = 36

where x = 5

that is 4×5 = 20

where y = 8

that is 2×8 = 16.

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The mean yearly rainfall in Sydney, Australia, is about 134 mm and the standard deviation is about 66 mm ("Annual maximums of,"

svetlana [45]

Answer:

At least 202.44 mm in the top 15%.

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

$\mu = 134, \sigma = 66$

How many yearly mm of rainfall would there be in the top 15%?

At least X mm.

X is the 100-15 = 85th percentile, which is X when Z has a pvalue of 0.85. So X when Z = 1.037.

$Z = \frac{X - \mu}{\sigma}$

$1.037 = \frac{X - 134}{66}$

$X - 134 = 66*1.037$

$X = 202.44$

At least 202.44 mm in the top 15%.

5 0

2 years ago

A salesman normally makes a sale (closes) on % of his presentations. Assuming the presentations are independent, find the pro

Tamiku [17]

Answer:

$(a)\ 0.0864$

$(b)\ 0.096$

$(c)\ 0.24$

$(d)\ 0.936$

Step-by-step explanation:

Given

$p \to$ close

$q \to$ fail to close

$p = 60\%$

$p = 0.60$

First, calculate the value of q

Using complement rule

$q = 1 - p$

$q = 1 - 0.60$

$q = 0.40$

So, we have:

$p = 0.60$ and $q = 0.40$

Solving (a): Fails to close on the 4th attempt

This means that he closes the first three attempts. The event is represented as: p p p q

So, we have:

$Pr = p*p*p*q$

$Pr = p^3*q$

$Pr = 0.60^3*0.40$

$Pr = 0.0864$

Solving (b): He closes for the first time on the 3rd attempt

This means that he fails to close the first two attempts. The event is represented as: q q p

So, we have:

$Pr = q * q * p$

$Pr = q^2 * p$

$Pr = 0.40^2 * 0.60$

$Pr = 0.096$

Solving (c): First he closes is his 2nd attempt

This means that he fails to close the first. The event is represented as: q p

So, we have:

$Pr = q * p$

$Pr = 0.40 * 0.60$

$Pr = 0.24$

Solving (d): The first he close is one of his 3 attempts

To do this, we make use of complement rule

The event that he does not close any of his first three attempts is: q q q

The probability is:

$Pr = q*q*q$

$Pr = q^3$

The opposite is that the first he closes is one of the first three

So, we have:

$Pr' = 1- Pr$ --- complement rule

$Pr' = 1- q^3$

$Pr' = 1- 0.40^3$

$Pr' = 1- 0.064$

$Pr' = 0.936$

4 0

2 years ago

At an archeological site, the remains of two ancient step pyramids are congruent. If ABCD~=EFGH, find AD

worty [1.4K]

Answer: 300ft

Step-by-step explanation:

3 0

2 years ago

Triangle ABC is a right triangle and cos(22.6o)=StartFraction b Over 13 EndFraction. Solve for b and round to the nearest whole

Art [367]

Answer:

a = 5 and b = 12

Step-by-step explanation:

<u>Step 1: Find angle B</u>

<em>Angle C = 90°</em>

<em>Angle A = 22.6°</em>

<em>Angle B = B</em>

<em>All angles in a triangle are equal to 180°.</em>

Angle A + Angle B + Angle C = 180°

22.6 + 90 + B = 180°

B = 180 - 112.6

B = 67.4°

<u>Step 2: Find the value of side AC 'b'</u>

<em>Hypotenuse = 13</em>

<em>Adjacent = b</em>

<em>Angle A = 22.6°</em>

Cos (Angle) = Adjacent/Hypotenuse

Cos (22.6) = b/13

b = 12

<u>Step 3: Find the value of side CB 'a'</u>

<em>Hypotenuse = 13</em>

<em>Opposite = a</em>

<em>Angle A = 22.6°</em>

Sin (angle) = Opposite/Hypotenuse

Sin (22.6°) = a/13

a = 4.99 rounded off to 5

Therefore, the value of a=5 and b=12.

!!

8 0

2 years ago

Read 2 more answers

Square $ABCD$ has area $200$. Point $E$ lies on side $\overline{BC}$. Points $F$ and $G$ are the midpoints of $\overline{AE}$ an

Charra [1.4K]

1. Consider square ABCD. You know that

$A_{ABCD}=AD^2=200,$

then

$AB=BC=CD=AD=10\sqrt{2}.$

2. Consider traiangle AED. F is mipoint of AE and G is midpoint of DE, then FG is midline of triangle AED. This means that

$FG=\dfrac{AD}{2}=\dfrac{10\sqrt{2} }{2}=5\sqrt{2}.$

3. Consider trapezoid BFGC. Its area is

$A_{BFGC}=\dfrac{FG+BC}{2}\cdot h,$ where h is the height of trapezoid and is equal to half of AB. Thus,

$A_{BFGC}=\dfrac{FG+BC}{2}\cdot \dfrac{AB}{2}=\dfrac{5\sqrt{2}+10\sqrt{2}}{2}\cdot \dfrac{10\sqrt{2}}{2}=75.$

4.

$A_{BFGC}=A_{BFGE}+A_{EGC},\\A_{EGC}=A_{BFGC}-A_{BFGE}=75-34=41.$

5. Note that angles EGC and CGD are supplementary and

$\sin \angle CGD=\sin \angle EGC.$

Then

$A_{CGD}=\dfrac{1}{2}CG\cdot CD\cdot \sin \angle CGD=\dfrac{1}{2}CG\cdot EG\cdot \sin \angle CGE=A_{ACG}=41.$

Answer: $A_{CGD}=41.$

8 0

2 years ago