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1 year ago

Can anyone help me with this question please? I want to finish it ASAP!

Mathematics

1 answer:

ivolga24 [154]1 year ago

5 0

Answer:

y = 2x + 1

Step-by-step explanation:

The gradient of line A is

$\frac{-6-(-2)}{0-2}=2$

Parallel lines have the same gradient, so m = 2.

Substituting the coordinates of P into y = mx + c,

$7=3(2)+c \implies c=1 \implies y=2x+1$

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Plz help .......................

Makovka662 [10]

$\frac{ 6\sqrt{2}}{ \sqrt{3} }$

Multiply the square root of 3 by the square root of 2 and itself.
The 3 cancels itself out and then you get A

6 0

3 years ago

Is there some one who can help me please

lesya692 [45]

Answer:

I'd say $18

Step-by-step explanation:

I cant really tell, but your best guess is $18

6 0

3 years ago

Find the range of 10 12 13 15 16.5

alex41 [277]

Answer:

15.5

Step-by-step explanation:

16.5-10

=15.5

So 15.5 is the range

6 0

3 years ago

A car traveling at 88 km an hour over takes a bus traveling at 64 km an hour if the bus has a 1.5 hour Headstart how far from th

PIT_PIT [208]

Answer: 352 miles

Step-by-step explanation:

At the point where the car overtook the bus, the car and the bus would have travelled the same distance.

Let x represent the distance covered by bus and the car before the car overtook the bus.

Let t represent the time taken by the bus to travel x miles.

The bus was traveling at 64 km an hour.

Distance = speed × time

Distance travelled by the bus in t hours would be

x = 64 × t = 64t

if the bus has a 1.5 hour Headstart, it means that the car started moving 1.5 hours after the bus has moved.

Therefore, time spent by the car would be t - 15 hours.

The car traveling at 88 km an hour

Distance covered by the car in t - 1.5 hours would be

x = 88(t - 1.5) = 88t - 132

Since the distance covered is equal, then

88t - 132 = 64t

88t - 64t = 132

24t = 132

t = 132/24

t = 5.5 hours

Therefore, the distance from the starting point when the car overtook the bus would be

x = 64 × 5.5 = 352 miles

3 0

3 years ago

The values of 60, 62, and 84 were common to both samples. The three values are identified as outliers with respect to the age-gr

postnew [5]

Answer:

Only 60 is an identified outlier

Step-by-step explanation:

Given

Age group: 40 to 50

$Min = 60$ $Q_1 = 70$ $Median= 73$

$Q_3 = 76$ $Max = 85$

Conditions for outlier

$Value > Q_3 + 1.5 * IQR$

$Value < Q_1 - 1.5 * IQR$

See attachment for complete question

Required

Which of 60, 62 and 84 is an outlier of age grout 40 to 50 years

First, calculate the IQR of group 40 to 50.

This is calculated as:

$IQR = Q_3 - Q_1$

Where:

$Q_3 = 76$ and $Q_1 = 70$

So:

$IQR = 76 - 70$

$IQR = 6$

Next, is to test the outlier conditions on each value (i.e. 60, 62 and 84)

Testing 60

Condition 1

$Value > Q_3 + 1.5 * IQR$

$60 > 76 + 1.5 * 6$

$60 > 76 + 9$

$60 > 85$ --- False

Condition 2

$Value < Q_1 - 1.5 * IQR$

$60 < 70 - 1.5 * 6$

$60 < 70 - 9$

$60 < 61$ --- True

Because one of the conditions is true, then 60 is an outlier of group 40 - 50 years

Testing 62

Condition 1

$Value > Q_3 + 1.5 * IQR$

$62> 76 + 1.5 * 6$

$62 > 76 + 9$

$62 > 85$ --- False

Condition 2

$Value < Q_1 - 1.5 * IQR$

$62 < 70 - 1.5 * 6$

$62 < 70 - 9$

$62 < 61$ --- False

Because both conditions are false, then 62 is not an outlier of group 40 - 50 years

Testing 84

Condition 1

$Value > Q_3 + 1.5 * IQR$

$84> 76 + 1.5 * 6$

$84 > 76 + 9$

$84 > 85$ --- False

Condition 2

$Value < Q_1 - 1.5 * IQR$

$84 < 70 - 1.5 * 6$

$84 < 70 - 9$

$84 < 61$ --- False

Because both conditions are false, then 84 is not an outlier of group 40 - 50 years

5 0

3 years ago