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2 years ago

7

jeff got a 23/25 on his first math test, a 93% on hi second test, and a 0.89 on his third test. Witch of the following orders hi

s scores from least to greatest?

1 answer:

vodka [1.7K]2 years ago

6 0

Answer:

0.89 23/25 93%

Step-by-step explanation:

hope this helps!!! :)

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Since Jordyn and Teri have fewer rolls of kite string, they decide to tie their string together so their kite can fly higher. Wr

torisob [31]

Question:

Jordyn, Teri, and Morgan are going to have a kite-flying contest. Jordyn and Teri each have one roll of kite string. They also each have 45 yards of extra string. Morgan has three rolls of kite string plus 10 yards of extra string. All the rolls of string are the same length. The girls want to see who can fly their kite the highest.

Since Jordyn and Teri have fewer rolls of kite string, they decide to tie their string together, so their kite can fly higher. Write at least two expressions to show how much kite string Jordyn and Teri have. Let x represent the number of yards of string on one roll.

Answer:

$y=x +45$

$y = 45 + x$

Step-by-step explanation:

From the complete question, we have:

<u>Jordyn & Teri</u>

<u></u> $Rolls = 1$ <u></u>

<u></u> $Extra = 45\ yd$ <u></u>

<u></u>

<u>Morgan</u>

<u></u> $Rolls = 3$ <u></u>

$Extra= 10\ yd$

Required

Expression for Jordyn and Morgan

Let the total kite by Jordyn and Teri be represented with y

This is calculated as:

$y = Rolls + Extra\ Yard$

This gives:

$y = 1\ roll + 45yd$

1 roll is represented with x.

So:

$y=1x +45$

$y=x +45$

This can be rewritten as:

$y = 45 + x$

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Kelly opened a bank account that earns 1.2% simple interest each year. After 7 years, Kelly will earn $126 in interest. How much

Alex

Answer:

13.4

Hope it helps that is what I think not that sure but I thing is corrwct!

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Find the sample size necessary in order to be 95% confident when determining the true mean weight within 2 units (EC2). Assume t

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Answer:

(c) 43.

Step-by-step explanation:

Given:

Margin of error (E) = 2

Sample variance $(s^{2})$ = 44

Confidence level = 95%

The z- value at 95% confidence level is 1.96.

The sample using the provided information can be calculated as:

$n=(\frac{z(s)}{E})^{2} \\n=\frac{z^{2} s^{2} }{E^{2} } \\n=\frac{1.96^{2} (44) }{2^{2} }\\n= 42.52\\n=43$

Hence, the correct option is (c) 43.

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3 years ago

What is the fractional part of a ton is a pound?

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