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3 years ago

Please help! :):):);)

Mathematics

1 answer:

AnnZ [28]3 years ago

6 0

Answer:

So 60% of this unknown number is 28

Let this unknown number be n

60/100 × n = 28

0.60 × n = 28

n = 28/0.6

n = 46.7

To the nearest wholw number is 47

So there are 47 votes in total

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kompoz [17]

Maybe you should ask your parents for some assistance with homework.

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3 years ago

A rubber ball is dropped onto a hard surface from a height of 9 feet, and it bounces up and down. At each bounce it rises to 80%

8_murik_8 [283]

There are 12 inches in a foot, so 9ft = 108in. Also, 80% = 0.8. Therefore the formula is: h(n) = 108 * 0.8^n. To find the bounce height after 10 bounces, substitute n=10 into the equation: h(n) = 108 * 0.8^10 = 11.60in (2.d.p.). Finally to find how many bounces happen before the height is less than one inch, substitute h(n) = 1, then rearrage with logarithms to solve for the power, x: 108 * 0.8^x = 1; 0.8^x = 1/108; Ln(0.8^x) = ln(1/108); xln(0.8) = ln(1\108); x = ln(1/108) / ln(0.8) = -4.682 / -0.223 = 21 bounces

4 0

3 years ago

Pls help! i don’t understand because i wasn’t paying attention in class :(

Anettt [7]

Answer:

1- $309.23

2- $107.10

3- 596.25

4- $8.18

There

Step-by-step explanation:

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4 0

3 years ago

An article presents a new method for timing traffic signals in heavily traveled intersections. The effectiveness of the new meth

Anna35 [415]

Answer:

With 89.9% we can say that the mean improvement is between 583.1 and 728.1 vehicles per hour.

Step-by-step explanation:

We are given that the effectiveness of the new method was evaluated in a simulation study. In 50 simulations, the mean improvement in traffic flow in a particular intersection was 655.6 vehicles per hour, with a standard deviation of 311.7 vehicles per hour.

A traffic engineer states that the mean improvement is between 583.1 and 728.1 vehicles per hour.

Let $\bar X$ = sample mean improvement

The z-score probability distribution for sample mean is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = mean improvement = 655.6 vehicles per hour

$\sigma$ = standard deviation = 311.7 vehicles per hour

n = sample of simulations = 50

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) 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.

Now, Probability that the mean improvement is between 583.1 and 728.1 vehicles per hour is given by = P(583.1 < $\bar X$ < 728.1) = P( $\bar X$ < 728.1) - P( $\bar X$ $\leq$ 583.1)

P( $\bar X$ < 728.1) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{728.1-655.6}{\frac{311.7}{\sqrt{50} } }$ ) = P(Z < 1.64) = 0.9495

P( $\bar X$ $\leq$ 583.1) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ $\leq$ $\frac{583.1-655.6}{\frac{311.7}{\sqrt{50} } }$ ) = P(Z $\leq$ -1.64) = 1 - P(Z < 1.64)

= 1 - 0.9495 = 0.0505

So, in the z table the P(Z $\leq$ x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 1.64 in the z table which has an area of 0.9495.

Therefore, P(583.1 < $\bar X$ < 728.1) = 0.9495 - 0.0505 = 0.899 or 89.9%

Hence, with 89.9% we can say that the mean improvement is between 583.1 and 728.1 vehicles per hour.

6 0

3 years ago

Please help! 3x – 4 = = 35

olganol [36]

Answer:

Step-by-step explanation:

3x13-4=35 Hope this helps

5 0

3 years ago

Read 2 more answers