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Nesterboy [21]
3 years ago
9

Willow has a recipe that asks her to use 3/4 cup of sugar. She is going to make 1/2 of the recipe, so she only needs 1/2 of that

3/4 cup. How many cups of sugar will she use?
Mathematics
1 answer:
Novosadov [1.4K]3 years ago
5 0

Answer:

3/8ths cup

Step-by-step explanation:

take 3/4 and times that by 2 then you get 6/8th and then take half of the 6 out and get 3/8th

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The first quartile can best be described as ______________________.
Zolol [24]

Answer:

A. the median of the lower half of a data set ordered from least to greatest

Step-by-step explanation:

The first quartile can best described as  the median of the lower half of a data set ordered from least to greatest.

Let us explain this with an example, say you're given a set of numbers: {1, 2, 5, 8, 9, 12, 15, 16, 20, 23, 25, 28, 32, 36, 42}. The median for the set is 16, the first quartile is going to be the median of the first half which is 8.

6 0
3 years ago
How do solve this equation step by step
Kryger [21]

Answer: 15n³-105n²+2n+16/6n²-42n

Step-by-step explanation:

n+8/3n²-21n +5n/2

2(n+8)+5n(3n²-21n)/2(3n²-21n)

2n+16+15n³-105n²/6n²-42n

15n³-105n²+2n+16/6n²-42n

7 0
3 years ago
Please help me please or i will fail my test please help
stich3 [128]

Answer:

1: B

Step-by-step explanation:

4 0
3 years ago
This problem has to do with Fahrenheit and Celsius
Vinvika [58]
F = 1.8C + 32

С = -10   ⇒   F = 1.8·(-10) + 32 = 14   ⇒   (-10, 14)
С = 0      ⇒   F = 1.8·(0) + 32 = 32      ⇒   (0, 32)
С = 10    ⇒   F = 1.8·(10) + 32 = 50    ⇒   (10, 50)
С = 20    ⇒   F = 1.8·(20) + 32 = 68    ⇒   (20, 68)

8 0
3 years ago
A study indicates that 37% of students have laptops. You randomly sample 30 students. Find the mean and the standard deviation o
Brilliant_brown [7]

Answer:

The mean and the standard deviation of the number of students with laptops are 1.11 and 0.836 respectively.

Step-by-step explanation:

Let <em>X</em> = number of students who have laptops.

The probability of a student having a laptop is, P (X) = <em>p</em> = 0.37.

A random sample of <em>n</em> = 30 students is selected.

The event of a student having a laptop is independent of the other students.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> and <em>p</em>.

The mean and standard deviation of a binomial random variable <em>X</em> are:

\mu=np\\\sigma=\sqrt{np(1-p)}

Compute the mean of the random variable <em>X</em> as follows:

\mu=np=30\times0.37=1.11

The mean of the random variable <em>X</em> is 1.11.

Compute the standard deviation of the random variable <em>X</em> as follows:

\sigma=\sqrt{np(1-p)}=\sqrt{30\times0.37\times(1-0.37)}=\sqrt{0.6993}=0.836

The standard deviation of the random variable <em>X</em> is 0.836.

5 0
3 years ago
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