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dsp73
3 years ago
7

A cereal company packages their granola in cylindrical containers that have a diamter of 20cm and a height of 17.4cm Approximate

ly how much granola will a container hold?
174cm cubed
348cm cubed
3140cm cubed
5464cm cubed
Mathematics
1 answer:
8_murik_8 [283]3 years ago
7 0

Answer:

the answer is 3,140 cm cubed

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Solve the Quadratics:
klasskru [66]

Answer:

Step-by-step explanation:

1 ) k²-8k = 0

k(k-8)=0

k = 0 or k=8

2) a²+5a=0

a(a+5) = 0

a=0 or a  = - 5

3 ) 6n²+5n-25=0

delta = b²-4ac      b =5 and a = 6   and   c= - 25

delta = 5²-4(6)(-25) = 625 = 25²

n 1 = (-5+25)/12 = 20/12 = 5/3

n 2 = (-5-25)/12 = -  30/12  = -5/2

same method for 4) and 5)

6 0
3 years ago
Gerry spends $1.25 each day on lunch at school. On Fridays, she buys an extra snack for $0.55. How much money will she spend in
Ludmilka [50]

Answer:

$13.60

Step-by-step explanation:

First let's find how much Gerry spends on lunch in 2 weeks (assuming she goes to school 5 days a week):

$1.25*10=$12.50

Now we find out how much she pays for the extra snack. Since in 2 weeks there are 2 Fridays then:

0.55*2=$1.10

Then we add these two together to get:

$12.50+$1.10=$13.60

4 0
2 years ago
Fill in the blank with the correct response.
Citrus2011 [14]

Answer:

remainder is 9

Step-by-step explanation:

3 0
3 years ago
Choose the correct simplification of the expression (−2x + 10y)(4x − 6y).
Korolek [52]
(-2x + 10y)(4x - 6y) =
-2x(4x - 6y) + 10y(4x - 6y) =
-8x^2 + 12xy + 40xy - 60y^2 =
-8x^2 + 52xy - 60y^2 <==
3 0
3 years ago
Read 2 more answers
The number of accidents per week at a hazardous intersection varies with mean 2.2 and standard deviation 1.4. The distribution o
DENIUS [597]

Answer:

33.36% probability that X is less than 2.

Step-by-step explanation:

The distribution is not normal, however, using the central limit theorem, it is going to be approximately normal. So

Central Limit Theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

Normal Probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 2.2, \sigma = 1.4, n = 9, s = \frac{1.4}{\sqrt{9}} = 0.467

(a) Suppose we let X be the mean number of accidents per week at the intersection during 9 randomly chosen weeks. What is the probability that X is less than 2?

This is the pvalue of Z when X = 2. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{2 - 2.2}{0.467}

Z = -0.43

Z = -0.43 has a pvalue of 0.3336.

33.36% probability that X is less than 2.

4 0
3 years ago
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