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1 year ago

Five cats each ate 1/4 cup of canned food . How much food did they eat altogether?

Mathematics

1 answer:

Sauron [17]1 year ago

4 0

We have FIVE cats that each ate 1/4 cup of canned food.

Then, the five of them together ate:

5 times 1/4 cup This in mathematical expression is a profduct of an integer value (5) times the fraction (1/4) which we know how to multiply based on the problems we solved a few minutes ago:

5 * 1/4 = 5/1 * 1/4 = (5 * 1) / (1 * 4) = 5/4 cups

This answer can also be written as a MIXED number given that 5/4 is an IMPROPER fraction (a fraction with numerator larger than the denominator)

5/4 = 1 1/4

Not sure what answer your teacher would prefer. It depends on whether he/she wants you to use fractions or mixed numbers.

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PLEASE HELP ASAP WILL MEDAL BRAINLIEST

Naddik [55]

Given
Present investment, P = 22000
APR, r = 0.0525
compounding time = 10 years
Future amount, A

A. compounded annually
n=10*1=10
i=r=0.0525
A=P(1+i)^n
=22000(1+0.0525)^10
=36698.11

B. compounded quarterly
n=10*4=40
i=r/4=0.0525/4
A=P(1+i)^n
=22000*(1+0.0525/4)^40
=37063.29

Therefore, by compounding quarterly, she will get, at the end of 10 years investment, an additional amount of
37063.29-36698.11
=$365.18

5 0

2 years ago

Wilhema bought 6 bars of soap for $12. The next day, Sophia bought 10 bars of the same kind of soap for $20. What is the cost of

kaheart [24]

Answer:

Step-by-step explanation:

To get this number you can divide the cost by the amount of bars of soap.

12/6=2

20/10=2

Hope this helps!

8 0

3 years ago

Which statement is true about the solution ?

Harrizon [31]

Answer:

There are no real solutions to this equation

Step-by-step explanation:

4 0

2 years ago

The mean MCAT score 29.5. Suppose that the Kaplan tutoring company obtains a sample of 40 students with a mean MCAT score of 32.

Paul [167]

Answer:

We conclude that the students that took the Kaplan tutoring have a mean score greater than 29.5.

Step-by-step explanation:

We are given that the Kaplan tutoring company obtains a sample of 40 students with a mean MCAT score of 32.2 with a standard deviation of 4.2.

Let $\mu$ = population mean score

So, Null Hypothesis, $H_0$ : $\mu \leq$ 29.5 {means that the students that took the Kaplan tutoring have a mean score less than or equal to 29.5}

Alternate Hypothesis, $H_A$ : $\mu$ > 29.5 {means that the students that took the Kaplan tutoring have a mean score greater than 29.5}

The test statistics that will be used here is One-sample t-test statistics because we don't know about population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean MCAT score = 32.2

s = sample standard deviation = 4.2

n = sample of students = 40

So, the test statistics = $\frac{32.2-29.5}{\frac{4.2}{\sqrt{40} } }$ ~ $t_3_9$

= 4.066

The value of t-test statistics is 4.066.

Now, at 0.05 level of significance, the t table gives a critical value of 1.685 at 39 degrees of freedom for the right-tailed test.

Since the value of our test statistics is more than the critical value of t as 4.066 > 1.685, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the students that took the Kaplan tutoring have a mean score greater than 29.5.

3 0

3 years ago

Solve question please

JulijaS [17]

Answer:

289

Step-by-step explanation:

$17^{2}$