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2 years ago

Isla walked 3/4 miles each way to and from the school Wednesday. How many miles did Isla walk that day?

Mathematics

2 answers:

attashe74 [19]2 years ago

8 0

1 1/2 is the answer 1 and 1 half

jolli1 [7]2 years ago

5 0

All you have to do is multiply 3/4 of a mile times two because she walked round trip. The answer is 1.5 miles

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If DE = 4x + 10, EF = 2x-1, and DF = 9x-15, find DF

Dmitrij [34]

Hello,
We have that:
(4x+10)+(2x-1)=9x-15

We solve the equation:
4x+10+2x-1=9x-15;
6x+9=9x-15;
6x-9x=-15-9;
-3x=-24;
3x=24;

from which
x=24:3=8

Then:
DF=9x-15=(9×8)-15=72-15=57

bye :-)

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2 years ago

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The equation r = 6t gives the number of albums released, r, by a record label over time in years, t. If you graph this relations

GalinKa [24]

R = 6t.....subbing in (8,48).....t = 8 and r = 48
48 = 6(8)
48 = 48 (correct)

r = 6t...subbing in (13,78)...t = 13 and r = 78
78 = 6(13)
78 = 78 (correct)

so u have 2 sets of points on this line and they are (8,48) and (13,78)

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2 years ago

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Vera_Pavlovna [14]

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1 year ago

Sixty-eight percent of adults in a certain country believe that life on other planets is plausible. You randomly select five adu

Pavlova-9 [17]

Answer:

Mean = 3.4

Variance = 1.088

Standard deviation = 1.0431

Step-by-step explanation:

$P=0.68\\ \\N=5$

The mean of a binomial distribution with parameters N (the number of trials) and p (the probability of success for each trial) is $m=N\cdot p$

Thus,

$m=5\cdot 0.68=3.4$

The variance of the binomial distribution is $s^2=Np(1-p)$ , where $s^2$ is the variance of the binomial distribution, so

$s^2=5\cdot 0.68\cdot (1-0.68)=3.4\cdot 0.32=1.088$

The standard deviation $s$ is the square root of the variance $s^2,$ so

$s=\sqrt{1.088}\approx 1.0431$

3 0

1 year ago

On average, indoor cats live to 16 years old with a standard deviation of 2.5 years. Suppose that the distribution is normal. Le

Y_Kistochka [10]

Answer:

(a) N (16, 2.5²)

(b) 0.241

High: 16.6 years

Step-by-step explanation:

The random variable X is defined as the age at death of a randomly selected indoor cat.

(a)

The distribution of X is:

$X\sim N(\mu = 16, \sigma^{2}=2.5^{2})$

(b)

Compute the probability that an indoor cat dies when it is between 17.2 and 19.6 years old as follows:

$P(17.2$

$=P(0.48$

Thus, the probability that an indoor cat dies when it is between 17.2 and 19.6 years old is 0.241.

(c)

Compute the two numbers within which 20% of indoor cats' age of death lies as follows:

$P(x_{1}$

The corresponding value of z is, 0.25.

Compute the value of x₁ and x₂ as follows:

$z=\frac{x_{1}-\mu}{\sigma}\\\\0.25=\frac{x_{1}-16}{2.5}\\\\x_{1}=16+(0.25\times 2.5}\\\\x_{1}=16.625\\\\x_{1}\approx 16.6$ $z=\frac{x_{1}-\mu}{\sigma}\\\\-0.25=\frac{x_{2}-16}{2.5}\\\\x_{2}=16-(0.25\times 2.5}\\\\x_{2}=15.375\\\\x_{2}\approx 15.4$

Low: 15.4 years

High: 16.6 years

8 0

2 years ago