Please someone help I don’t understand this

Simplify the following expression [ (3x6) ] + (5x2) ] ÷ 7

KonstantinChe [14]

[ (3x6) ] + (5x2) ] ÷ 7
= (18 + 10)/7
= 28/7
= 4

6 0

3 years ago

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Math question picture included A.40m B. 36m C. 32 m D. 30 m

Katarina [22]

Answer: 32 meters

Explanation: The 4 meter pole casts a shadow of 5 meters. The towers unknown height casts a shadow of 40 meters. Now isolate the shadows. 5 and 40. 40 divided by 5 is 8, so 4 times 8 is 32.

6 0

3 years ago

Write an equation of the perpendicular bisector of the segment with end points M(1,5) and N(7,-1)

vitfil [10]

The perpendicular bisector of the segment passes through the midpoint of this segment. Thus, we will initially find the midpoint P:

$P=\dfrac{(1,5)+(7,-1)}{2}=\dfrac{(8,4)}{2}=(4,2)$

Now, we will calculate the slope of the segment support line (r). After this, we will use the fact that the perpendicular bisector (p) is perpendicular to r:

$m_r=\dfrac{\Delta y}{\Delta x}=\dfrac{5-(-1)}{1-7}=\dfrac{6}{-6}\iff m_r=-1$

$p\perp r\Longrightarrow m_p\cdot m_r=-1\Longrightarrow m_p\cdot(-1)=-1\iff m_p=1$

We can calculate the equation of p by using its slope and its point P:

$y-y_P=m_p(x-x_P)\\\\
y-2=1\cdot(x-4)\\\\
y-2=x-4\\\\
\boxed{p:~~y=x-2}$

4 0

3 years ago

What is the domain of the function f(x)=6x+7

Lostsunrise [7]

The domain is all real numbers, reason being, you can put any number for x and get a valid coordinate.

5 0

3 years ago

Researchers measured the data speeds for a particular smartphone carrier at 50 airports. The highest speed measured was 75.6 Mbp

olga55 [171]

Answer:

a) 59.98

b) 2.99

c) 2.99

d) Significantly High

Step-by-step explanation:

Part a)

Highest speed measured = x = 75.6 Mbps

Average/Mean speed = $\overline{x}$ = 15.62 Mbps

Standard Deviation = s = 20.03 Mbps

We need to find the difference between carrier's highest data speed and the mean of all 50 data speeds i.e. x - $\overline{x}$

x - $\overline{x}$ = 75.6 - 15.62 = 59.98 Mbps

Thus, the difference between carrier's highest data speed and the mean of all 50 data speeds is 59.98 Mbps

Part b)

In order to find how many standard deviations away is the difference found in previous part, we divide the difference by the value of standard deviation i.e.

$\frac{59.98}{20.03}=2.99$

This means, the difference is 2.99 standard deviations or in other words we can say, the Carrier's highest data speed is 2.99 standard deviations above the mean data speed.

Part c)

A z score tells us that how many standard deviations away is a value from the mean. We calculated the same in the previous part. Performing the same calculation in one step:

The formula for the z score is:

$z=\frac{x-\overline{x}}{s}$

Using the given values, we get:

$z=\frac{75.6-15.62}{20.03}=2.99$

Thus, the Carriers highest data is equivalent to a z score of 2.99

Part d)

The range of z scores which are neither significantly low nor significantly high is -2 to + 2. The z scores outside this range will be significant.

Since, the z score for carrier's highest data speed is 2.99 which is well outside the given range, i.e. greater than 2, we can conclude that the carrier's highest data speed is significantly higher.

3 0

3 years ago