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2 years ago

On a map of Texas, the

Mathematics

1 answer:

sergejj [24]2 years ago

6 0

1 inch on the map = 50 miles on the Earth.

A certain trip on the map is 2-3/4 inches.

-- first inch = 50 miles

-- second inch = another 50 miles

-- 3/4 inch = 3/4 of 50 miles (37.5 miles)

Total:

Here's an equation;

1 map-inch = 50 real-miles

Multiply each side by 2-3/4 :

2-3/4 inches = (2-3/4) x (50 miles)

2-3/4 map-inches = 137.5 real-miles

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What percent is the change in the price if: a The price was $100 and now it is $1250?

KatRina [158]

Answer:

The increase is 1150.

11.5 times the original value, or 1150% of it. That's a big increase, and I would not buy that thing, whatever it is.

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3 years ago

The number of views in a cat and a dog after they uploaded are represented by the following

alexandr1967 [171]

Answer:

i see nothin

Step-by-step explanation:

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2 years ago

QUESTION 8. An unknown number is added to 8. The sum is

Rudik [331]

Answer:

Step-by-step explanation:

Let the unknown number be x

If an unknown number is added to 8, this is expressed as;

8 + x

If The sum is then divided by 9 to give the answer 10, then;

(8+x)/9 = 10

Cross multiply

8 + x = 9 * 10

8 + x = 90

x = 90 - 8

x = 82

Hence the unknown number is 82

3 0

2 years ago

Solve for t. -18 = 6t A. t = -3 B. t = -1/3 C. t = 12 D. t = 24

dangina [55]

Answer:

A) t = -3

Step-by-step explanation:

-18 = 6t

divide both sides of the equation by 6

-18/6 = 6t/6

-18/6 = t

-3 = t

6 0

2 years ago

Read 2 more answers

An average computer mouse inspector can inspect 50 mice per hour. The 48 computer mice inspectors at a particular factory can on

Ronch [10]

Answer: C. Yes, because -2.77 falls in the critical region .

Step-by-step explanation:

Let $\mu$ be the population mean .

As per given , we have

$H_0:\mu=50\\\\ H_a: \mu$

Since the alternative hypothesis is left-tailed and population standard deviation is not given , so we need to perform a left-tailed t-test.

Test statistic : $t=\dfrac{\overline{x}-\mu}{\dfrac{s}{\sqrt{n}}}$

Also, it is given that ,

n= 48

$\overline{x}=46$

s= 10

$t=\dfrac{46-50}{\dfrac{10}{\sqrt{48}}}=\dfrac{-4}{\dfrac{10}{6.93}}\\\\=\dfrac{-4}{1.443}\approx-2.77$

Degree of freedom = df = n-1= 47

Using t-distribution , we have

Critical value = $t_{\alpha,df}=t_{0.025,47}=2.0117$

Since, the absolute t-value (|-2.77|=2.77) is greater than the critical value.

So , we reject the null hypothesis.

i.e. -2.77 falls in the critical region.

[Critical region is the region of values that associates with the rejection of the null hypothesis at a given probability level.]

Conclusion : We have sufficient evidence to support the claim that these inspectors are slower than average.

Hence, the correct answer is C. Yes, because -2.77 falls in the critical region

4 0

3 years ago