All categories

3 years ago

Need help with this question.

Mathematics

1 answer:

Vaselesa [24]3 years ago

4 0

Answer:

4:1 (I think)

Step-by-step explanation:

Let's write some equations first:

For both prices increasing by 20:

$\frac{x+20}{y+20} =\frac{5}{2}$

then we can use cross multiplication

2x+40=5y+100

For both prices decreasing by 5:

$\frac{x-5}{y-5} =\frac{5}{1}$

same here:

5y-25=x-5

Using substitution for 5y, we get: 5y=x+20

2x+40=5y+100

2x+40=x+20+100

x=80

plugging this in to the second equation:

5y-25=x-5

5y-25=80-5

5y-25=75

5y=100

y=20

therefore we get x:y=80:20

4:1

You might be interested in

Find the perimeter of the given image. Round your answers to the<br> nearest tenth.

Paha777 [63]

Answer:

52.6 mm

Step-by-step explanation:

split this figure and form one square and semi circle

for semicircle

radius = 4mm

find diameter

radius = diameter /2

4*2 = diameter

8 mm = diameter

now find perimeter of square

POS = 4l

=4*8

=32 mm

perimeter of semicircle= πr + 2r

=3.14^4 + 2*4

=12.56 + 8

=20.56 mm

area of the figure = perimeter of square + perimeter of semicircle

=32 + 20.56 mm

=52.56 mm

=52.6 mm

split ur figure as shown below

4 0

3 years ago

Read 2 more answers

Chloe swam 40 laps in the pool, but this was only 50% of her total swimming workout. How many more laps does she still need to s

Alja [10]

She has to 40 more laps. 50% is a half, so 40 is half of something. so you are dividing 40 by a half, which is 80. half of 80 is 40.

6 0

3 years ago

Ursula has 1/2 cup of butter. She uses 1/4 cup of butter to make a batch of brownies. How many batches of brownies can Ursula ma

Oksana_A [137]

Ursula can make 2 batches of brownies

8 0

4 years ago

Which point is a solution to the inequality in this graph

olganol [36]

Given:

The graph of an inequality.

To find:

The point which is a solution of the given graph of inequality.

Solution:

From the given graph it is clear that the boundary line of the graph is a dotted line. It means the points lie in shaded region are in the solution set but the points on the line are not included in the solution set.

The points (3,2) and (-3,-6) are lie on the boundary line. it means they are not the solution of the inequality represented by the given graph.

Point (5,0) lies on the positive x-axis at the distance of 5 units from the origin and it doesn't lies in the shaded region. So, (5,0) is not a solution.

Point (0,5) lies on the negative y-axis at the distance of 5 units from the origin and it lies in the shaded region. So, (0,5) is a solution.

Therefore, the correct option is B.

8 0

3 years ago

Items are inspected for flaws by two quality inspectors. Both inspectors inspect every item and the probability that an item has

Genrish500 [490]

Answer:

a)0.976

b)0.00926

c)0.2402

d)0.35

Step-by-step explanation:

Let $X_i$ be an item passed by inspector i

Let Y be the event that there is a fault in an item

The probability that an item has a flaw is 0.1 i.e. P(Y)=0.1

If a flaw is present ,it will be detected by the first inspector with probability 0.92 i.e. $P(\bar{X_1}|Y)=0.92$

So, $P(X_1|Y)=1-0.92=0.08$

If a flaw is present ,it will be detected by the second inspector with probability 0.7 i.e. $P(\bar{X_2}|Y)=0.7$

So, $P(X_2|Y)=1-0.7=0.3$

If an item does not have a flaw, it will be passed by the first inspector with probability 0.95 i.e. $P(X_1|\bar{Y}) = 0.95$

So, $P(\bar{X_1}|\bar{Y}) = 1-0.95=0.05$

If an item does not have a flaw, it will be passed by the second inspector with probability 0.8 i.e. $P(X_2|\bar{Y}) = 0.8$

So, $P(\bar{X_2}|\bar{Y}) = 1-0.8=0.2$

a)P(found by atleast one inspector | It has flaw )=1-P(found by none inspector | It has flow )

P(found by atleast one inspector | It has flaw )= $1-P(X_1|Y) P(X_2|Y)$

P(found by atleast one inspector | It has flaw )= $1-0.08 \times 0.3$

P(found by atleast one inspector | It has flaw )=0.976

Hence the probability that it will be found by at least one of the two inspectors if it has flaw is 0.976

b) $P(Y|X_1)=\frac{P(X_1|Y) P(Y)}{P(X_1|Y) P(Y)+P(X_1|\bar{Y}) P(\bar{Y})}$

$P(Y|X_1)=\frac{0.08 \times 0.1}{0.08 \times 0.1+0.95 \times 0.9}=0.00926$

C)P( two inspectors draw different conclusions on the same item)= $P(X_1 \cap \bar{X_2} \cap Y)+P(\bar{X_1} \cap X_2 \cap Y)+P(X_1 \cap \bar{X_2} \cap \bar{Y})+P(\bar{X_1} \cap X_2 \cap \bar{Y})$

P( two inspectors draw different conclusions on the same item)=0.2402

$P(Y|(X_1 \cap X_2))=\frac{P(Y \cap X_1 \cap X_2)}{P(X_1 \cap X_2)}\\P(Y|(X_1 \cap X_2))=\frac{P(Y \cap X_1 \cap X_2)}{P(X_1 \cap X_2 \cap Y)+P(X_1 \cap X_2 \cap \bar{Y})}\\P(Y|(X_1 \cap X_2))=0.35$

3 0

3 years ago