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zzz [600]
3 years ago
9

PLEASE HELP ASAP!!!!!!!!!! I AM IN 7TH GRADE MATH FOR FLVS AND I NEED HELP WHOEVER HELPS ME THE FASTEST GETS BRAINIEST!!!!!!!!!!

!!!!
I already did the chart just need help with this one question

1. What is the total cost of the entire bill (all four meals), including tip? Please show work!

Mathematics
2 answers:
coldgirl [10]3 years ago
6 0
The total is 157.2, all you had to do was add all the numbers under the column "Total Bill"
mylen [45]3 years ago
6 0
34.80+78+16.80+27.60=157.2
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The center of a circle is at (-7, -2) and the radius is 7. What is the equation of the circle in standard form?
Liono4ka [1.6K]

Answer:

<h2>(x + 7)² + (y + 2)² = 49</h2>

Step-by-step explanation:

The standard form of an equation of a circle:

(x-h)^2+(y-k)^2=r^2

(h, k) - center

r - radius

We have:

center (-7, -2) → h = -7 and k = -2

radius r = 7.

Substitute:

(x-(-7))^2+(y-(-2))^2=7^2\\\\(x+7)^2+(y+2)^2=49

8 0
3 years ago
The average daily volume of a computer stock in 2011 was μ=35.1 million​ shares, according to a reliable source. A stock analyst
alexandr1967 [171]

Answer:

The 95% confidence interval is ( 27.126 , 34.674)

Step-by-step explanation:

Given

The t critical value at 0.05 level = 2.023 for the df = 39

Confidence interval = 95%

Mean  

\bar{x} - t * \frac{s}{\sqrt{n} } < \mu < \bar{x} + t * \frac{s}{\sqrt{n} }

Substituting the given values we get -

\30.9 - 2.023 * \frac{11.8}{\sqrt{40} } < \mu < \30.9 + 2.023 * \frac{11.8}{\sqrt{40} }\\27.126 < \mu < 34.674

Hence, the 95% confidence interval is

( 27.126 , 34.674)

4 0
3 years ago
The radius of circle O is 15 m. What is the length of the diameter?
satela [25.4K]

Answer:

30m

Step-by-step explanation:

diameter= radius times 2

d=(15)(2)

d=30

8 0
3 years ago
Read 2 more answers
Hey you click and answer :}
Contact [7]

Answer:

Hey :)

Step-by-step explanation:

3 0
3 years ago
Find the probability for the experiment of tossing a coin three times. Use the sample space S = {HHH, HHT, HTH, HTT, THH, THT, T
myrzilka [38]

Answer:

1) 0.375

2) 0.375

3) 0.5

4) 0.5

5) 0.875

6) 0.5                          

Step-by-step explanation:

We are given the following in the question:

Sample space, S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}.

\text{Probability} = \displaystyle\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}

1. The probability of getting exactly one tail

P(Exactly one tail)

Favorable outcomes ={HHT, HTH, THH}

\text{P(Exactly one tail)} = \dfrac{3}{8} = 0.375

2. The probability of getting exactly two tails

P(Exactly two tail)

Favorable outcomes ={ HTT,THT, TTH}

\text{P(Exactly two tail)} = \dfrac{3}{8} = 0.375

3. The probability of getting a head on the first toss

P(head on the first toss)

Favorable outcomes ={HHH, HHT, HTH, HTT}

\text{P(head on the first toss)} = \dfrac{4}{8} = \dfrac{1}{2} = 0.5

4. The probability of getting a tail on the last toss

P(tail on the last toss)

Favorable outcomes ={HHT,HTT,THT,TTT}

\text{P(tail on the last toss)} = \dfrac{4}{8} = \dfrac{1}{2} = 0.5

5. The probability of getting at least one head

P(at least one head)

Favorable outcomes ={HHH, HHT, HTH, HTT, THH, THT, TTH}

\text{P(at least one head)} = \dfrac{7}{8} = 0.875

6. The probability of getting at least two heads

P(Exactly one tail)

Favorable outcomes ={HHH, HHT, HTH,THH}

\text{P(Exactly one tail)} = \dfrac{4}{8} = \dfrac{1}{2} = 0.5

3 0
3 years ago
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