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3 years ago

7

How much money is saved on an $80 coat that has been reduced 30%. If you could show me how this is done I would be super gratefu

l

1 answer:

pentagon [3]3 years ago

6 0

You would get 24$ off of it

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Alejandro needs to buy bird seed from the local pet store for his parrots. A

Gnesinka [82]

First take note he only has $20 and the bag he must buy is $12.55. Also the loose seeds are $1.49 per pound.

To solve subtract 12.55 from 20

20-12.55=7.45

So after buying the 10 lb bag he has $7.45 left now divide that by 1.49

7.45/1.49=5

So after buying the 10 lb bag he can also buy 5 lb of the loose seeds!

7 0

3 years ago

The admission fee at an amusement park is $1.50 for children and $4 for adults. On a certain day 333 people enter the park, the

liberstina [14]

Answer:

<u>188 children</u> and <u>145 adults</u> were admitted in the park.

Step-by-step explanation:

Given:

The admission fee at an amusement park is $1.50 for children and $4 for adults.

Total $862 collected on a certain day when 333 people enter the park.

Now, to find the children and adults admitted in the park.

<u><em>Let the number of children admitted be </em></u> $x.$ <u><em /></u>

<u><em>And the the number of adults admitted be </em></u> $y.$ <u><em /></u>

So, the total people enter the park:

$x+y=333\\\\x=333-y\ \ \ ...(1)$

Thus, the total amount collected of the admission fee:

$1.50(x)+4(y)=862\\\\$

Substituting the value of $x$ from equation (1):

$1.50(333-y)+4(y)=862\\\\499.50-1.50y+4y=862\\\\499.50+2.50y=862\\\\Subtracting\ both\ sides\ by\ 499.50\ we\ get:\\\\2.50y=362.50\\\\Dividing\ both\ sides\ by\ 2.50\ we\ get:\\\\y=145.$

<u><em>Thus, the number of adults = 145.</em></u>

Now, to get the number of children by substituting the value of $y$ in equation (1) we get:

$x=333-y\\\\x=333-145\\\\x=188.$

<u><em>Hence, the number of children = 188.</em></u>

Therefore, 188 children and 145 adults were admitted in the park.

4 0

3 years ago

In ΔUVW, \overline{UW}

kozerog [31]

Answer: is extended through point W to point X angle

Explanation

8 0

2 years ago

HELLLLLLLLLLLLLLPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP

MrRa [10]

Answer:

96

Step-by-step explanation:

you have to find the answer to 8t by the power of 2 multiplied witch is 64 + 32t = 96

7 0

2 years ago

One stats class consists of 52 women and 28 men. Assume the average exam score on Exam 1 was 74 (σ = 10.43; assume the whole cla

Svetllana [295]

Answer:

(A) What is the z- score of the sample mean?

The z- score of the sample mean is 0.0959

(B) Is this sample significantly different from the population?

No; at 0.05 alpha level (95% confidence) and (n-1 =79) degrees of freedom, the sample mean is NOT significantly different from the population mean.

Step -by- step explanation:

(A) To find the z- score of the sample mean,

X = 75 which is the raw score

¶ = 74 which is the population mean

S. D. = 10.43 which is the population standard deviation of/from the mean

Z = [X-¶] ÷ S. D.

Z = [75-74] ÷ 10.43 = 0.0959

Hence, the sample raw score of 75 is only 0.0959 standard deviations from the population mean. [This is close to the population mean value].

(B) To test for whether this sample is significantly different from the population, use the One Sample T- test. This parametric test compares the sample mean to the given population mean.

The estimated standard error of the mean is s/√n

S. E. = 16/√80 = 16/8.94 = 1.789

The Absolute (Calculated) t value is now: [75-74] ÷ 1.789 = 1 ÷ 1.789 = 0.559

Setting up the hypotheses,

Null hypothesis: Sample is not significantly different from population

Alternative hypothesis: Sample is significantly different from population

Having gotten T- cal, T- tab is found thus:

The Critical (Table) t value is found using

- a specific alpha or confidence level

- (n - 1) degrees of freedom; where n is the total number of observations or items in the population

- the standard t- distribution table

Alpha level = 0.05

1 - (0.05 ÷ 2) = 0.975

Checking the column of 0.975 on the t table and tracing it down to the row with 79 degrees of freedom;

The critical t value is 1.990

Since T- cal < T- tab (0.559 < 1.990), refute the alternative hypothesis and accept the null hypothesis.

Hence, with 95% confidence, it is derived that the sample is not significantly different from the population.

6 0

3 years ago