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Valentin [98]
3 years ago
6

A roast beef sandwich cost $6.75. A customer buys multiple sandwiches. Write and equation using x to represent the number of san

dwiches
Mathematics
2 answers:
GREYUIT [131]3 years ago
4 0

Answer:

y = 6.75x

Step-by-step explanation:

Shalnov [3]3 years ago
4 0

Answer:

Step-by-step explanation:

She gives 4 pumpkins

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Simplify (8x^2 - 1 + 2x^3) - (7x^3 - 3x^2 + 1).
Elenna [48]

Answer:

31x^2-2 is the answer....

Step-by-step explanation:

7 0
3 years ago
HCF of 405 and 1605 answer it fast
Rashid [163]

Answer:

hcf=(1,605; 600) = 3 × 5

Step-by-step explanation:

Prime Factorization of a number: finding the prime numbers that multiply together to make that number.

1,605 = 3 × 5 × 107;

1,605 is not a prime, is a composite number;

600 = 23 × 3 × 52;

600 is not a prime, is a composite number;

* Positive integers that are only dividing by themselves and 1 are called prime numbers. A prime number has only two factors: 1 and itself.

* A composite number is a positive integer that has at least one factor (divisor) other than 1 and itself.

Multiply all the common prime factors, by the lowest exponents (if any).

gcf, hcf, gcd (1,605; 600) = 3 × 5

gcf, hcf, gcd (1,605; 600) = 3 × 5 = 15;

The numbers have common prime factors.

Must mark brainliest for more answers.

And you should friend me.

7 0
3 years ago
The answer is down below please help meeee
monitta

Answer:

TheaterMania, basically you are getting more popcorn for $1, thats how did it.

7 0
3 years ago
Read 2 more answers
A random sample of 10 parking meters in a beach community showed the following incomes for a day. Assume the incomes are normall
Vlad1618 [11]

Answer: (2.54,6.86)

Step-by-step explanation:

Given : A random sample of 10 parking meters in a beach community showed the following incomes for a day.

We assume the incomes are normally distributed.

Mean income : \mu=\dfrac{\sum^{10}_{i=1}x_i}{n}=\dfrac{47}{10}=4.7

Standard deviation : \sigma=\sqrt{\dfrac{\sum^{10}_{i=1}{(x_i-\mu)^2}}{n}}

=\sqrt{\dfrac{(1.1)^2+(0.2)^2+(1.9)^2+(1.6)^2+(2.1)^2+(0.5)^2+(2.05)^2+(0.45)^2+(3.3)^2+(1.7)^2}{10}}

=\dfrac{30.265}{10}=3.0265

The confidence interval for the population mean (for sample size <30) is given by :-

\mu\ \pm t_{n-1, \alpha/2}\times\dfrac{\sigma}{\sqrt{n}}

Given significance level : \alpha=1-0.95=0.05

Critical value : t_{n-1,\alpha/2}=t_{9,0.025}=2.262

We assume that the population is normally distributed.

Now, the 95% confidence interval for the true mean will be :-

4.7\ \pm\ 2.262\times\dfrac{3.0265}{\sqrt{10}} \\\\\approx4.7\pm2.16=(4.7-2.16\ ,\ 4.7+2.16)=(2.54,\ 6.86)

Hence, 95% confidence interval for the true mean= (2.54,6.86)

7 0
4 years ago
What is the answer to -26-(-49)?
Radda [10]

Answer:

23

Step-by-step explanation:

-26-(-49)=-26+49=23

7 0
3 years ago
Read 2 more answers
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