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-Dominant- [34]
2 years ago
14

The best drink at the Smoothie Shack is the Berry Blaster. It takes 3 cups of strawberries to make 2 Berry Blasters. The Smoothi

e Shack only has 15 cups of strawberries left.
How many Berry Blasters can the Smoothie Shack make?
Mathematics
1 answer:
Bess [88]2 years ago
4 0

Answer:

The Smoothie Shack can make 10 Berry Blaster

Step-by-step explanation:

Hope this helps or right at least. :^

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Each of the letters below represents a different digit. if EAT = 721 what does TURKEY represent
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The data set represents the total number of people who bought bananas each hour at a grocery store.
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Suppose that the ages of members in a large billiards league have a known standard deviation of σ = 12 σ=12sigma, equals, 12 yea
asambeis [7]

Answer:

n\geq 23

Step-by-step explanation:

-For a known standard deviation, the sample size for a desired margin of error is calculated using the formula:

n\geq (\frac{z\sigma}{ME})^2

Where:

  • \sigma is the standard deviation
  • ME is the desired margin of error.

We substitute our given values to calculate the sample size:

n\geq (\frac{z\sigma}{ME})^2\\\\\geq (\frac{1.96\times 12}{5})^2\\\\\geq 22.13\approx23

Hence, the smallest desired sample size is 23

3 0
2 years ago
*NO 194 39% 13:08 Pilih jawaban yang paling tepat Segitiga PQR dan segitiga XYZ 5 poin merupakan dua segitiga yang kongruen. Bes
Illusion [34]

Answer:

Option d. PQ = YZ

Step-by-step explanation:

<u><em>The question in English is</em></u>

Choose the most appropriate answer. The PQR triangle and the XYZ triangle are two congruent triangles. The angle P = the angle Y and PR = YX. Side pairs of the same length are ... a. PQ = XZ b. QR = YZ c. QR = XY d. PQ = YZ

we know that

If two triangles are congruent, then its corresponding sides and its corresponding angles are congruent

Corresponding sides are named using pairs of letters in the same position on either side of the congruence statement

so

we have that

m\angle P=m\angle Y

PR=YX

so

Triangle PQR is congruent with Triangle YZX

That means

<em><u>Corresponding angles</u></em>

m\angle P=m\angle Y

m\angle Q=m\angle Z

m\angle R=m\angle X

<u><em>Corresponding sides</em></u>

PQ=YZ\\QR=ZX\\PR=YX

8 0
2 years ago
Find the smallest relation containing the relation {(1, 2), (1, 4), (3, 3), (4, 1)} that is:
professor190 [17]

Answer:

Remember, if B is a set, R is a relation in B and a is related with b (aRb or (a,b))

1. R is reflexive if for each element a∈B, aRa.

2. R is symmetric if satisfies that if aRb then bRa.

3. R is transitive if satisfies that if aRb and bRc then aRc.

Then, our set B is \{1,2,3,4\}.

a) We need to find a relation R reflexive and transitive that contain the relation R1=\{(1, 2), (1, 4), (3, 3), (4, 1)\}

Then, we need:

1. That 1R1, 2R2, 3R3, 4R4 to the relation be reflexive and,

2. Observe that

  • 1R4 and 4R1, then 1 must be related with itself.
  • 4R1 and 1R4, then 4 must be related with itself.
  • 4R1 and 1R2, then 4 must be related with 2.

Therefore \{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(4,1),(4,2)\} is the smallest relation containing the relation R1.

b) We need a new relation symmetric and transitive, then

  • since 1R2, then 2 must be related with 1.
  • since 1R4, 4 must be related with 1.

and the analysis for be transitive is the same that we did in a).

Observe that

  • 1R2 and 2R1, then 1 must be related with itself.
  • 4R1 and 1R4, then 4 must be related with itself.
  • 2R1 and 1R4, then 2 must be related with 4.
  • 4R1 and 1R2, then 4 must be related with 2.
  • 2R4 and 4R2, then 2 must be related with itself

Therefore, the smallest relation containing R1 that is symmetric and transitive is

\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(2,1),(2,4),(3,3),(4,1),(4,2),(4,4)\}

c) We need a new relation reflexive, symmetric and transitive containing R1.

For be reflexive

  • 1 must be related with 1,
  • 2 must be related with 2,
  • 3 must be related with 3,
  • 4 must be related with 4

For be symmetric

  • since 1R2, 2 must be related with 1,
  • since 1R4, 4 must be related with 1.

For be transitive

  • Since 4R1 and 1R2, 4 must be related with 2,
  • since 2R1 and 1R4, 2 must be related with 4.

Then, the smallest relation reflexive, symmetric and transitive containing R1 is

\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(2,1),(2,4),(3,3),(4,1),(4,2),(4,4)\}

5 0
2 years ago
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