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3 years ago

She counted the money and found out she has $14.90 in the jar. How many dimes and quarters does she have?

Mathematics

1 answer:

nikdorinn [45]3 years ago

5 0

Answer:

There're too much possibilities of the answer.

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A small country is reducing its government spending. The country's annual debt is modeled by y = 500(1 − 0.40)t, where y is in b

lesya692 [45]

D because it just makes sense

3 0

3 years ago

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Ivan has 4 hours of Music on his MP3 player jazz makes up 1/6 of the total time how many hours of jazz are on Ivan's MP3 player

Alex777 [14]

4 hrs = 240 min.

Jazz = 1/6(240) = 40 min.

Therefore, Ivan's MP3 player has 40 min (2/3 hr) of jazz music.

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3 years ago

Rick took his family out for dinner. He planned to leave a 15% gratuity on the bill. What is the total cost of the bill is 123.5

Mazyrski [523]

Answer:

142.03 would be your answer

Step-by-step explanation:

You can multiply 123.50 by 15, which gets you $1852.50. Divide that by 100 and you are left with $18.525, which you can round to $18.53. Add that to what the cost of the bill was and you are left with $142.03

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4 years ago

A student wanted to estimate the number of chocolate chips in a commercial brand of cookie. He sampled 100 cookies and found an

m_a_m_a [10]

Answer: A. (8.4,12.6)

Step-by-step explanation:

Confidence interval $(\mu)$ is given by :-

$\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}$

, where n is the sample size

$\sigma$ = Population standard deviation.

$\overline{x}$ = Sample mean

$z_{\alpha/2}$ = Two tailed z-value for significance level of $\alpha$ .

Given : Confidence level = 99% = 0.99

Significance level = $\alpha=1-0.99=0.01$

By standard normal z-value table ,

Two tailed z-value for Significance level of 0.01 :

$z_{\alpha/2}=z_{0.005}=2.576$

Also,

n=100

$\sigma= 8$

$\overline{x}=10.5$

Then, the required 99% confidence interval for the average number $(\mu)$ of chips per cookie :-

$10.5\pm (2.576)\dfrac{8}{\sqrt{100}}\\\\ =10.5\pm 2.0608\\\\=(10.5-2.0608,\ 10.5+2.0608)=(8.4392,\ 12.5608)\approx(8.4,\ 12.6)$

Hence, the 99% confidence interval for the average number of chips per cookie = (8.4,12.6)

5 0

3 years ago

What number is 45% of 5/9

igor_vitrenko [27]

The right answer is 1/4

7 0

3 years ago

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