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3 years ago

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the weight M of an object on Mars varies directly as it’s weight E on Earth. A person who weighs 95 lb on Earth weighs 38 lb on

Mars. How much would a 100 lb person weight on Mars

1 answer:

lorasvet [3.4K]3 years ago

8 0

Answer:

A 100 lb person will weight 40 lb on Mars

Step-by-step explanation:

Given:

Weight on Mars directly varies as its weight on earth.

Let weight on Mars be = $M$

Let weight on Earth be = $E$

Its known that:

$M$ ∝ $E$

Thus

$M=kE$

where $k$ is constant of proportionality.

when $E=95\ lb$ , then $M=38\ lb$

thus, we have

$38=k(95)$

dividing both sides by 95.

$\frac{38}{95}=\frac{k(95)}{95}$

∴ $k=\frac{38}{95}$

Thus when $E= 100$ , $M$ would be calculated as

$M=\frac{38}{95}\times 100$

$M=\frac{3800}{95}$

∴ $M=40\ lb$ (Answer)

∴ A 100 lb person will weight 40 lb on Mars.

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Help pls. this topic is confusing for me

Leya [2.2K]

Answer:

My brother took this test and said it was 35. Im not sure but try it

Step-by-step explanation:

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3 years ago

A piece of fabric shaped as a trapezoid has a

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Answer:

Area of the fabric = 175 square inches

Step-by-step explanation:

Area of a trapezoid = (b1 + b2)/2 * h

Where,

b1 = 16 inches

b2 = 9 inches

h = 14 inches

Area of a trapezoid = (b1 + b2)/2 * h

= (16 + 9)/2 * 14

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3 years ago

In a study of the accuracy of fast food drive-through orders, McDonald’s had 33 orders that were not accurate among 362 orders o

melomori [17]

Answer:

A. We need to conduct a hypothesis in order to test the claim that the true proportion of inaccurate orders p is 0.1.

B. Null hypothesis: $p=0.1$

Alternative hypothesis: $p \neq 0.1$

C. $z=\frac{0.0912 -0.1}{\sqrt{\frac{0.1(1-0.1)}{362}}}=-0.558$

D. $z_{\alpha/2}=-1.96$ $z_{1-\alpha/2}=1.96$

E. Fail to the reject the null hypothesis

F. So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the true proportion of inaccurate orders is not significantly different from 0.1.

Step-by-step explanation:

Data given and notation

n=362 represent the random sample taken

X=33 represent the number of orders not accurate

$\hat p=\frac{33}{363}=0.0912$ estimated proportion of orders not accurate

$p_o=0.10$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

A: Write the claim as a mathematical statement involving the population proportion p

We need to conduct a hypothesis in order to test the claim that the true proportion of inaccurate orders p is 0.1.

B: State the null (H0) and alternative (H1) hypotheses

Null hypothesis: $p=0.1$

Alternative hypothesis: $p \neq 0.1$

When we conduct a proportion test we need to use the z statistic, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

C: Find the test statistic

Since we have all the info required we can replace in formula (1) like this:

$z=\frac{0.0912 -0.1}{\sqrt{\frac{0.1(1-0.1)}{362}}}=-0.558$

D: Find the critical value(s)

Since is a bilateral test we have two critical values. We need to look on the normal standard distribution a quantile that accumulates 0.025 of the area on each tail. And for this case we have:

$z_{\alpha/2}=-1.96$ $z_{1-\alpha/2}=1.96$

P value

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a bilateral test the p value would be:

$p_v =2*P(z$

E: Would you Reject or Fail to Reject the null (H0) hypothesis.

Fail to the reject the null hypothesis

F: Write the conclusion of the test.

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the true proportion of inaccurate orders is not significantly different from 0.1.

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3 years ago

Identify the postulate that proves the triangles are congruent.

Ratling [72]

Answer:

ASA

Step-by-step explanation:

Triangle ABD is congruent to triangle BDC by ASA because angle B in ABD is 30 degrees. Then they give you BD which equals BD by the reflexive property. And finally angle D in ABD is also 30 degrees. In triangle BCD, angle B is 30 degrees, BD = BD, and angle D in BCD is 30 degress. So we have an angle (B), side (BD), and another angle (D)

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3 years ago

Nitric acid (specific gravity 1.42) costs $51.95 per liter. A student spills 5.112 kg of nitric acid in the laboratory, what is

Leno4ka [110]

Answer:

The cost of wasted nitric acid is $187.02

Step-by-step explanation:

Since it is given that specific gravity of Nitric Acid is 1.42 Thus the density of nitric acid is

$\rho _{acid}=S.G\times \rho _{water}\\\\\\\rho_{acid}=1.42\times 1\frac{g}{cm^3}\\\\\rho_{acid}=1.42g/cm^3$

Now from the basic relation of mass volume and density we have

$\rho =\frac{Mass}{Volume}\\\\\therefore Volume=\frac{Mass}{\rho }$

Since the mass of the nitric acid is given to be 5.112 kg = 5112 grams

Thus the volume of nitric acid is

$Volume=\frac{5112}{1.42}=3600cm^3$

Now we know that 1000 cubic centimeters euals 1 litrer

Thus the calculated volume in liters is

$V_{liters}=\frac{3600}{1000}=3.6Liters$

Thus the cost of acid spilled equals

$Cost=3.6\times 51.95=187.02$

5 0

3 years ago