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andrew-mc [135]

2 years ago

5

An international company has 19,700 employees in one country. If this represents 29.4% of the company's employees, how many empl

oyees does it have in total?
Round your answer to the nearest whole number. Please helP I will mark brainliest

1 answer:

koban [17]2 years ago

7 0

Answer:

I believe it is 490530 or 4905.3

Step-by-step explanation:

I used a calculator for this

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Help please please help

Semenov [28]

Answer:

Step-by-step explanation:

1,8

5,7

2,3

3 0

3 years ago

Read 2 more answers

Sandra owns a rectangular pice of farmland that is 2/3 miles wide and has an area 5/9 square mile what is the leght of sandras f

Rzqust [24]

Answer:

5/6 mile

Step-by-step explanation:

The formula for the area of a rectangle can be used with the given values to write an equation for the length of the farmland.

__

Here is the equation for the area of a rectangle.

A = LW . . . . . area is the product of length and width

When we fill in the area and width, we have ...

5/9 mi² = L(2/3 mi) . . . . . equation for the length of the farmland

Solving this equation gives ...

L = (5/9)/(2/3) mi . . . . . . . . divide by the coefficient of L

L = (5/9)/(6/9) mi = 5/6 mi . . . . . perform the division

The length of Sandra's farmland is 5/6 mile.

6 0

1 year ago

Write the explicit formula for this geometric sequence.

Naily [24]

The explicit formula for this geometric sequence 81, 27, 9, 3, … is aₙ = 81(1/3)ⁿ⁻¹. It has common ratio of 1/3 and first term of 81.

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

An equation is an expression that shows the relationship between two or more variables and numbers.

A geometric sequence is in the form aₙ = ar⁽ⁿ⁻¹⁾, where a is the first term and r is the common ratio.

The geometric sequence 81, 27, 9, 3, … has common ratio:

r = 27/81 = 1/3

The explicit formula for this geometric sequence 81, 27, 9, 3, … is aₙ = 81(1/3)ⁿ⁻¹. It has common ratio of 1/3 and first term of 81.

Find out more on equation at: brainly.com/question/2972832

#SPJ1

7 0

1 year ago

Solution 2^2x+3-7(2^2x+1)+3=0 introduce Log

Andrej [43]

Answer:

$x = \frac{log\sqrt{-1/6}}{log2}$

Step-by-step explanation:

Given the expression

$2^{2x}+3-7(2^{2x}+1)+3=0$

Let $P=2^x$

Substituting into the expression, we will have:

$P^2+3-7(P^2+1)+3=0\\Expand\\P^2+3-7P^2-7+3=0\\-6P^2-1=0\\6P^2=-1\\p^2=-1/6\\P=\sqrt{-1/6}$

Since:

$P=2^x\\2^x=\sqrt{-1/6}\\xlog2=log(\sqrt{-1/6}) \\x = \frac{log\sqrt{-1/6}}{log2}$

7 0

2 years ago

Unlike most packaged food products, alcohol beverage container labels are not required to show calorie or nutrient content. An a

zhenek [66]

Answer:

We conclude that the true average estimated calorie content in the population sampled exceeds the actual content.

Step-by-step explanation:

We are given that an article reported on a pilot study in which each of 58 individuals in a sample was asked to estimate the calorie content of a 12 oz can of beer known to contain 153 calories.

The resulting sample mean estimated calorie level was 193 and the sample standard deviation was 88.

Let $\mu$ = <u><em>true average estimated calorie content in the population sampled.</em></u>

So, Null Hypothesis, $H_0$ : $\mu$ $\leq$ 153 calories {means that the true average estimated calorie content in the population sampled does not exceeds the actual content}

Alternate Hypothesis, $H_A$ : $\mu$ > 153 calories {means that the true average estimated calorie content in the population sampled exceeds the actual content}

The test statistics that would be used here <u>One-sample t test statistics</u> as we don't know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean estimated calorie level = 193 calories

s = sample standard deviation = 88

n = sample of individuals = 58

So, <u><em>the test statistics</em></u> = $\frac{193-153}{\frac{88}{\sqrt{58} } }$ ~ $t_5_7$

= 3.462

The value of t test statistics is 3.462.

Since, in the question we are not given the level of significance so we assume it to be 5%. <u>Now, at 0.05 significance level the t table gives critical value of 1.6725 at 57 degree of freedom for right-tailed test.</u>

Since our test statistic is more than the critical value of t as 3.462 > 1.6725, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <u>we reject our null hypothesis</u>.

Therefore, we conclude that the true average estimated calorie content in the population sampled exceeds the actual content.

8 0

2 years ago