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anastassius [24]
2 years ago
12

Edger needs 6 cookies and 2 brownies for every 4 plates how many cookies and brownies does he need for 10 plates

Mathematics
1 answer:
evablogger [386]2 years ago
6 0

Answer:

Edger needs 15 cookies and 5 brownies.

Step-By-Step:

6 + 6 + 3 = 15

2 + 2 + 1 = 5

4 can go into 10, 2.5 times

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A linear model is given for the data in the table: y=1.25x+2.
Fed [463]

Answer:

Option A

Step-by-step explanation:

Given that A linear model is given for the data in the table: y=1.25x+2.

Let us write observed values for each x and also the predicted values as per equation.

x            2     3        4      8    10      16    20    24    Total

y((O)      3      4        7     12   16      22   28    30

y(P)     4.5   5.75      7     12   14.5   22   27    32

DEv    1.5    1.75      0       0     1.5     0      1       2     7 75

where y(0) represents observed y or y in the table given

y(P) gives values of y predicted as per the equation 1.25x+2

Dev represents the absolute difference

Hence answer is option

A.7.75

8 0
2 years ago
100+100 omg this is so tricky ooooo​
amm1812

Answer:

200

Step-by-step explanation:

100+

100

-----

200

8 0
3 years ago
Read 2 more answers
Multiply. Enter your answer as a simplified mixed number. -4(3/14) ​
Rus_ich [418]

Answer:

-53 /14

Step-by-step explanation:

6 0
2 years ago
Please help i will give brainliest!!!
Shalnov [3]
D) (-2,3)

-2 is the x-value
3 is the y-value

If you look at the graph on the x-axis (horizontal line) the numbers are increasing and decreasing by 1.
Since the point of intersection (big blue dot) is on the left side of 0, it falls directly at the -2 mark
As for the y-axis (vertical lines) the numbers also increase and decrease by 1. The point of intersection falls above 0 at +3

Therefore your point of intersection is (-2,3)
8 0
2 years ago
A company has a policy of retiring company cars; this policy looks at number of miles driven, purpose of trips, style of car and
pav-90 [236]

Answer:

ans=13.59%

Step-by-step explanation:

The 68-95-99.7 rule states that, when X is an observation from a random bell-shaped (normally distributed) value with mean \mu and standard deviation \sigma, we have these following probabilities

Pr(\mu - \sigma \leq X \leq \mu + \sigma) = 0.6827

Pr(\mu - 2\sigma \leq X \leq \mu + 2\sigma) = 0.9545

Pr(\mu - 3\sigma \leq X \leq \mu + 3\sigma) = 0.9973

In our problem, we have that:

The distribution of the number of months in service for the fleet of cars is bell-shaped and has a mean of 53 months and a standard deviation of 11 months

So \mu = 53, \sigma = 11

So:

Pr(53-11 \leq X \leq 53+11) = 0.6827

Pr(53 - 22 \leq X \leq 53 + 22) = 0.9545

Pr(53 - 33 \leq X \leq 53 + 33) = 0.9973

-----------

Pr(42 \leq X \leq 64) = 0.6827

Pr(31 \leq X \leq 75) = 0.9545

Pr(20 \leq X \leq 86) = 0.9973

-----

What is the approximate percentage of cars that remain in service between 64 and 75 months?

Between 64 and 75 minutes is between one and two standard deviations above the mean.

We have Pr(31 \leq X \leq 75) = 0.9545 = 0.9545 subtracted by Pr(42 \leq X \leq 64) = 0.6827 is the percentage of cars that remain in service between one and two standard deviation, both above and below the mean.

To find just the percentage above the mean, we divide this value by 2

So:

P = {0.9545 - 0.6827}{2} = 0.1359

The approximate percentage of cars that remain in service between 64 and 75 months is 13.59%.

4 0
3 years ago
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