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3 years ago

The difference between 4,632 and 20,000 is what number?

2 answers:

4 0

We find the difference of the two numbers by subtracting the two numbers given above. It is calculated as follows:

Difference = 20000 - 4632 = 15368

Therefore, the difference between 4,632 and 20,000 is 15368.

Hope this answers the question. Have a nice day.

slavikrds [6]3 years ago

3 0

We have to find the difference between these two numbers by subtracting.

20,000 - 4,632 = 15,368

Answer = the difference between these two numbers is 15,368.

Good Luck! :)

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Find the value of x in the triangle shown below. D is x=8

SVEN [57.7K]

Step-by-step explanation:

${2}^{2} + {x}^{2} = {6}^{2} \\ 4 + {x}^{2} = 36 \\ {x}^{2} = 36 - 4 \\ {x}^{2} = 32 \\ x = \sqrt{32}$

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3 years ago

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3 years ago

The dependent variable, y is 6 less than the independent variable, x.

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D) because “y is” means “y equals” leaving only 2 choices, and it said “6 less than the independent variable, 6. (Y=6-x)

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3 years ago

How can I solve (b)?

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90/3 parts of silver = 30 times multiplied by
tin = 12 × 30 = 360g
lead = 8 × 30 = 240g

12.5 / 23 = 0.5434 parts per metal

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3 years ago

A local car dealer claims that 25% of all cars in San Francisco are blue. You take a random sample of 600 cars in San Francisco

sammy [17]

Answer:

No, we can't reject the dealer's claim with a significance level of 0.05.

Step-by-step explanation:

We are given that a local car dealer claims that 25% of all cars in San Francisco are blue.

You take a random sample of 600 cars in San Francisco and find that 141 are blue.

Let p = proportion of all cars in San Francisco who are blue

SO, Null Hypothesis, $H_0$ : p = 25% {means that 25% of all cars in San Francisco are blue}

Alternate Hypothesis, $H_A$ : p $\neq$ 25% {means that % of all cars in San Francisco who are blue is different from 25%}

The test statistics that will be used here is One-sample z proportion statistics;

T.S. = $\frac{\hat p-p}{{\sqrt{\frac{\hat p(1-\hat p)}{n} } } } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of 600 cars in San Francisco who are blue = $\frac{141}{600}$ = 0.235

n = sample of cars = 600

So, test statistics = $\frac{0.235-0.25}{{\sqrt{\frac{0.235(1-0.235)}{600} } } } }$

= -0.866

Now at 0.05 significance level, the z table gives critical values of -1.96 and 1.96 for two-tailed test. Since our test statistics lies within the range of critical values of z so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that 25% of all cars in San Francisco are blue which means the dealer's claim was correct.

4 0

3 years ago