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raketka [301]
3 years ago
8

Mr Murray’s distance away (in meters) from his classroom after t seconds can be modeled by the following equation: M(t) = 3.5t +

2. After 3 seconds, how far is Mr Murray away from his classroom?
Mathematics
1 answer:
Vaselesa [24]3 years ago
5 0
Substitute the given value of three into the equation so that it becomes

M= 3.5*3+2

M= 12.5

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John goes to a restaurant and has a bill of $32.57. He uses a 10% off coupon on
otez555 [7]

Answer:

<u>John spends $ 37.98</u>

Step-by-step explanation:

1. John has a bill of $ 32.57

2. He leaves a tip of 18% on the amount before  the coupon or tax is applied, in consequence: 32.57 + 18% (32.57) = 32.57 + 5.86 = 38.43

3. He uses a 10% off coupon on  the cost of the meal, therefore:

38.43 - 10% (32.57) = 38.43 - 3.26 = 35.17

4. The tax is 8%, then:

35.17 + 8% (35.17) = 35.17 + 2.81 = 37.98

<u>John spends $ 37.98</u>

5 0
3 years ago
HELP PLS HELP PLS
ivanzaharov [21]

Answer:

x =-4

y=-1

x=2

y=8

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5 0
3 years ago
G
Svetllana [295]

Answer:

15

Step-by-step explanation:

since there is 7 teams and 12 can be the minimum we can do 7 x 12=84

this means that 3 people could possibly be on one single team.

the maximum number is 12+3=15

5 0
2 years ago
Let us suppose we have data on the absorbency of paper towels that were produced by two different manufacturing processes. From
maksim [4K]

Answer:

The 95% CI for the difference of means is:

-155.45 \leq \mu_1-\mu_2 \leq -44.55

Step-by-step explanation:

<em>The question is incomplete:</em>

<em>"Find a 95% confidence interval on the difference of the towels mean absorbency produced by the two processes. Assumed that the standard deviations are estimated from the data. Round to two decimals places."</em>

Process 1:

- Sample size: 10

- Mean: 200

- S.D.: 15

Process 2:

- Sample size:  4

- Mean: 300

- S.D.: 50

The difference of the sample means is:

M_d=M_1-M_2=200-300=-100

The standard deviation can be estimated as:

\sigma_d=\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}\\\\\sigma_d=\sqrt{\frac{15^2}{10}+\frac{50^2}{4}} =\sqrt{22.5+625}=\sqrt{647.5}=25.45

The degrees of freedom are:

df=n_1+n_2-2=10+4-2=12

The t-value for a 95% confidence interval and 12 degrees of freedom is t=±2.179.

Then, the confidence interval can be written as:

M_d-t\cdot \sigma_d\leq \mu_1-\mu_2 \leq M_d+t\cdot \sigma_d\\\\-100-2.179*25.45\leq \mu_1-\mu_2 \leq -100+2.179*25.45\\\\-100-55.45 \leq \mu_1-\mu_2 \leq -100+55.45\\\\ -155.45 \leq \mu_1-\mu_2 \leq -44.55

8 0
3 years ago
Find the distance between the points (2, 4) and (8,-8) on a coordinate plane, to the nearest whole number.
Alisiya [41]

Answer:

2,5

Step-by-step explanation:

2 to the right and 5 up

8 0
3 years ago
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