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2 years ago

On a map scale, I inch = 53 miles. Two attractions are eight inches apart on the map. How far are they apart in miles?

Mathematics

1 answer:

soldi70 [24.7K]2 years ago

5 0

Answer:

424 miles

Step-by-step explanation:

53 x 8 = 424 miles

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Zolol [24]

The number of the specimens would be in the culture after 5 minutes will be 7,962.62.

<h3>What is an exponent?</h3>

Consider the function:

y = a (1 + r) ˣ

Where x is the number of times this growth occurs, a = initial amount, and r = fraction by which this growth occurs.

A bacteria culture grows at a rate of 20% per minute.

If the culture initially has 3,200 bacteria specimens.

Then the equation will be

y = 3200 (1 + 0.2) ˣ

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Then the number of the specimens would be in the culture after 5 minutes will be

y = 3200 (1.20)⁵

y = 7962.62

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1 year ago

If the test for a disease is accurate 82% of the time, how often will it come back negative if a patient has the disease?

aniked [119]

Would it just be 100 - 82 = 18, so D maybe?

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2 years ago

Read 2 more answers

A cake manufacturer boxes Swiss chocolate and German chocolate cakes at one site. A box of Swiss chocolate cake contains 0.55 lb

Kobotan [32]

Answer:

2,500 German chocolate cake boxes.

1,500 Swiss chocolate cake boxes.

Step-by-step explanation:

Let 'S' be the number of Swiss chocolate cakes boxed and 'G' the number of German cholocate cakes boxed. If all of the available ingredients are used:

$0.55*S +0.59*G = 2,300\\0.45*S+0.41*G = 1,700$

Solving the linear system above:

$(0.55 *S +0.59*G)- \frac{0.55}{0.45} (0.45*S+0.41*G)= 2,300-\frac{0.55}{0.45}*1,700*\\0.088888*G = 222.222\\G=2,500\\S =\frac{2,300-2,500*0.59}{0.55}\\S=1500$

2,500 German chocolate cake boxes and 1,500 Swiss chocolate cake boxes can be made each day.

6 0

3 years ago

Please help

Alekssandra [29.7K]

Answer:

25 in x 15 in

Step-by-step explanation:

Given:

Length = 3/5 the width
Area = 375 in²

Let width = $x$

Therefore, length = 3/5 $x$

First create an equation for the area of the picture based on the given information for its width and length:

$\begin{aligned} \implies \textsf{Area of original picture} & = \sf width \times length\\& = x\left(\dfrac{3}{5}x\right)\\& = \dfrac{3}{5}x^2\end{aligned}$

We are told the area of the enlarged picture is 375 in². Therefore, substitute this into the equation and solve for $x$ to find the width of the enlarged picture:

$\begin{aligned}\textsf{Area} & = 375\\ \implies \dfrac{3}{5}x^2 & = 375\\ x^2 & =375 \cdot \dfrac{5}{3}\\ x^2 & =625\\ x & = \sqrt{625}\\ x& = 25\end{aligned}$

Therefore, the width of the enlarged picture is 25 in.

Substitute the found value of $x$ into the expression for length to find the length of the enlarged picture:

$\begin{aligned}\sf Length & = \dfrac{3}{5}x\\\\\implies \sf Length & = \dfrac{3}{5}(25)\\\\& = 15\: \sf in\end{aligned}$

Therefore, the dimensions of the enlarged picture are <u>25 in x 15 in</u>. The width is 25 in and the length is 15 in, as the length is 3/5 of the width.

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1 year ago

Find 1/6 of an hour?

maks197457 [2]

Answer:

10 minuites

Step-by-step explanation:

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2 years ago

On a map scale, I inch = 53 miles. Two attractions are eight inches apart on the map. How far are they apart in miles?​

On a map scale, I inch = 53 miles. Two attractions are eight inches apart on the map. How far are they apart in miles?