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Anarel [89]
2 years ago
12

You are in a bike race. When you get to the first checkpoint, you are $\frac{2}{5}$

Mathematics
1 answer:
Black_prince [1.1K]2 years ago
6 0

Answer:

The distance from the star to the second checkpoint is 4 miles

Step-by-step explanation:

The complete question in the attached figure

we know that

When you get to the second checkpoint, you are 1/4 of the distance to the finish

so

The distance from the star to the second checkpoint is equal to the total distance multiplied by 1/4

The total distance is 40 miles ( see the attached figure)

so

The distance from the star to the second checkpoint is

\frac{1}{4}(40)=10\ miles

<em>Find out the distance from the start to the first checkpoint</em>

Multiply the distance from the star to the second checkpoint by 2/5

\frac{2}{5}(10)=4\ miles

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A quality control officer randomly checking weights of pumpkin seed bags being filled by an automatic filling machine. Each bag
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From 0g to 494.7g are rejected as well as those heavier than 505.3grams.

You calculate this by subtracting and adding 5.3 to 500
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1 year ago
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La distancia de Urano al Sol es 2 870 990 000 km y la distancia de la Tierra al Sol es 1,496 × 108 km. Aproximadamente, ¿cuántas
Leni [432]

Answer:

2,7204x 10^9

Step-by-step explanati

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6 0
2 years ago
The diagram shows a logo made from three circles
ratelena [41]

Answer:

Use the formula π*(r^2) where r is radius

Area of big circle, all 3=314.1592654 (approximately) and this is =100%

Area of middle circle=153.93804

Area of small circle=78.53981634

Percentage of middle circle with small circle = (153.93804/314.1592654)*100= 48.999999999999 approx= 49%

Percentage of small circle alone = (78.53981634/314.1592654)*100

= 25%

So 51%= big circle alone

And 51%+25%= 76%

100%-76%=24%

3 0
2 years ago
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A given set of values is found to be a normal distribution with a mean of 140 and a standard deviation of 18.0. Find the value t
Alisiya [41]

Answer:

The value that is greater than 45% of the data values is approximately 137.84.

Step-by-step explanation:

The key is transforming values from this distribution to a z-score range and finding the corresponding value using a z-score table.

We are looking for a value x which attains a critical z-score that corresponds to the (100-45)%=55-th percentile:

z_{0.55} = \frac{x-\mu}{\sigma}=\frac{x-140}{18}\implies x = 18\cdot z_{0.55}+140

The critical z value (from z-score table, online) is: -0.12, so:

x = 18\cdot z_{0.55}+140=18\cdot(-0.12)+140\approx137.84

The value that is greater than 45% of the data values is approximately 137.84.


5 0
2 years ago
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The cubit is an ancient unit. Its length equals six palms. (A palm varies from 2.5 to 3.5 inches depending on the individual.) W
olasank [31]

Assuminhg the ark has a shoe-box (cuboid) shape it´s volume would be:

V_{cuboid}=lenght*width*height

We have this three measurements (l=300cubits;w=50cubits;h=30cubits)) and it can be as simple as replacing them in the equation and solving but they are in all in cubits. We will convert them to ft because the problem requires the answer in ft^{3}. In order to do this we will use the given equivalences:

1cubit=6palms\\1palm=3.10in

and another one:

1ft=12in

First we will convert from cubits to palms:

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now from palms to in:

l=1800palms*\frac{3.10in}{1palm}=5580in\\w=300palms*\frac{3.10in}{1palm}=930in\\h=180palms*\frac{3.10in}{1palm}=558in\\

now from in to ft:

l=5580in*\frac{1ft}{12in}=465ft\\w=930in*\frac{1ft}{12in}=77.5ft\\h=558in*\frac{1ft}{12in}=46.5ft

We can calculate the Volume now like this:

V_{ark}=465ft*77.5ft*46.5ft=1675743.75ft^{3}

The volume of trhe ark would be 1675743.75ft^{3}

6 0
1 year ago
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