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Aloiza [94]
2 years ago
13

What is the percent change from 70 to 56

Mathematics
1 answer:
Dmitry_Shevchenko [17]2 years ago
3 0
\frac{(56 - 70)}{|70|}  × 100

\frac{-14}{70} × 100 

-2.0 × 100 → 20%

Therefore, your "percent change" is  → 20% 
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A car dealership sells SUVs and passenger cars . For a recent year, 80 more SUVs were sold than passenger cars. If a total of 51
MAVERICK [17]

Answer:

295 SUVs were sold, and 215 passenger vehicles.

Step-by-step explanation:

Let s and p represent the number of SUVs and p the number of passenger cars.  

Then s = p + 80 and s + p = 510.

Substitute p + 80 for s in the second equation, obtaining:

p + 80 + p = 510, or:

2p = 430

Dividing both sides by 2 isolates p:  p = 215.

Since s = p + 80, we add 80 to 215 here, obtaining s = 295.

295 SUVs were sold, and 215 passenger vehicles.

3 0
3 years ago
To make ribbon and bow for a hat, Stacey needs 5/6 yard of black ribbon and 2/3 yard of red ribbon. how much total ribbon does s
larisa86 [58]
First make the denominators of the fractions the same by multiplying.

\frac{2}{3} × \frac{2}{2}
=\frac{4}{6}

Now we just need to add the fractions together

\frac{5}{6} +  \frac{4}{6}
= \frac{9}{6}

9/6 would be the answer, but it can be simplified further, then converted into decimal form.

\frac{9}{6} ÷ \frac{3}{3}
=\frac{3}{2}
=1.5 yards

Your final answer is 1.5 yards of ribbon are needed.
7 0
3 years ago
Read 2 more answers
An octagonal swimming pool has a base area of 22 square feet The pool is 3 feet deep How many cubic feet of water can the pool h
LenKa [72]

The pool can hold 65.84 ft³ of water

<u>Explanation:</u>

Given:

Shape of pool = octagonal

Base area of the pool = 22 ft²

Depth of the pool = 3 feet

Volume, V = ?

We know:

Area of octagon = 2 ( 1 + √2) a²

22 ft² = 2 ( 1 + √2 ) a²

\frac{11}{1+\sqrt{2} } = a^2

a² = \frac{11}{2.42}

a² = 4.55

a = 2.132 ft

Side length of the octagon is 2.132 ft

We know:

Volume of octagon = 2(1+\sqrt{2} ) X (a)^2 X h

V = 2(1+\sqrt{2})X (2.132)^2 X 3\\ \\V = 2 ( 2.414) X 4.5454 X 3\\\\V = 65.84 ft^3

Therefore, the pool can hold 65.84 ft³ of water

8 0
3 years ago
Concerns about the climate change and CO2 reduction have initiated the commercial production of blends of biodiesel (e.g. from r
natta225 [31]

Answer:

a) 99% of the sample means will fall between 0.933 and 0.941.

b) By the Central Limit Theorem, approximately normal, with mean 0.937 and standard deviation 0.0015.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be 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.

Central Limit Theorem

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

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

(a) If the true mean is 0.9370 with a standard deviation of 0.0090 within what interval will 99% of the sample means fail?

Samples of 34 means that n = 34

We have that \mu = 0.937, \sigma = 0.009

By the Central Limit Theorem, s = \frac{0.009}{\sqrt{34}} = 0.0015

Within what interval will 99% of the sample means fail?

Between the (100-99)/2 = 0.5th percentile and the (100+99)/2 = 99.5th percentile.

0.5th percentile:

X when Z has a pvalue of 0.005. So X when Z = -2.575.

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

By the Central Limit Theorem

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

-2.575 = \frac{X - 0.937}{0.0015}

X - 0.937 = -2.575*0.0015

X = 0.933

99.5th percentile:

X when Z has a pvalue of 0.995. So X when Z = 2.575.

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

2.575 = \frac{X - 0.937}{0.0015}

X - 0.937 = 2.575*0.0015

X = 0.941

99% of the sample means will fall between 0.933 and 0.941.

(b) If the true mean 0.9370 with a standard deviation of 0.0090, what is the sampling distribution of ¯X?

By the Central Limit Theorem, approximately normal, with mean 0.937 and standard deviation 0.0015.

6 0
3 years ago
EXPERTSSSS HELP (q.5)
Ostrovityanka [42]

Answer:

not paying 20% means that she DID pay %80

Step-by-step explanation:

4 0
2 years ago
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