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ExtremeBDS [4]
3 years ago
10

How long is a dinner fork in inches.

Mathematics
2 answers:
Juliette [100K]3 years ago
4 0
About 4 inches long

Hope this helps
Alja [10]3 years ago
4 0
4 in but 4 to 5 if it is dinner but it depends  if its baby of kid or adult  

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What is the LCM of 504 and 594.
Elza [17]

Answer: The LCM is 16632.

Step-by-step explanation: please..just divide 16632 and 504 and 594 or something...

I’m to tired to give an explanation just take my word for it.

Now I’m gonna take a long nap after that..

Hope this helps!

6 0
3 years ago
Which is bigger -7/11 or -2/3
Bad White [126]

Answer:

-2/3

Step-by-step explanation:

3 0
3 years ago
I don’t understand any of this
AysviL [449]

Answer:

Step-by-step explanation:

From the given triangles in the picture attached,

∠A ≅ ∠X [Given]

∠B ≅ ∠Y [Given]

By AA property of similarity of the triangles, both the triangles will be similar.

ΔABC ~ ΔXZY

Similarity statement:

Corresponding sides of these similar triangles will be proportional.

\frac{XZ}{AB}= \frac{ZY}{BC}=\frac{XY}{AC}

Similarity ratio:

\frac{XY}{AC}=\frac{24}{16}

      = \frac{3}{2}

      = 3 : 2

5 0
3 years ago
Please help I do not know how to do this
nordsb [41]

Answer:

About 33% of students prefer chocolate icecream

Step-by-step explanation:

The actual percentage would be 35%, but none of the other answers are close enough to correct, so I'd have to say the bottom one.

hope this helps you :/

3 0
3 years ago
A random sample of 28 statistics tutorials was selected from the past 5 years and the percentage of students absent from each on
SVEN [57.7K]

Answer:

Step-by-step explanation:

Hello!

X: number of absences per tutorial per student over the past 5 years(percentage)

X≈N(μ;σ²)

You have to construct a 90% to estimate the population mean of the percentage of absences per tutorial of the students over the past 5 years.

The formula for the CI is:

X[bar] ± Z_{1-\alpha /2} * \frac{S}{\sqrt{n} }

⇒ The population standard deviation is unknown and since the distribution is approximate, I'll use the estimation of the standard deviation in place of the population parameter.

Number of Absences 13.9 16.4 12.3 13.2 8.4 4.4 10.3 8.8 4.8 10.9 15.9 9.7 4.5 11.5 5.7 10.8 9.7 8.2 10.3 12.2 10.6 16.2 15.2 1.7 11.7 11.9 10.0 12.4

X[bar]= 10.41

S= 3.71

Z_{1-\alpha /2}= Z_{0.95}= 1.645

[10.41±1.645*(\frac{3.71}{\sqrt{28} } )]

[9.26; 11.56]

Using a confidence level of 90% you'd expect that the interval [9.26; 11.56]% contains the value of the population mean of the percentage of absences per tutorial of the students over the past 5 years.

I hope this helps!

7 0
4 years ago
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