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Jet001 [13]
3 years ago
15

Raymond spends 1/10 of his vacation budget on a car rental. Write an expression for how much Raymond has to spend on the rest of

his vacation. If he budgets $650 for the vacation, how much will he have left to spend?
A. (1/10) b, $65
B. (9/10) b, $65
C. (9/10) b, $585
D. (1/10) b, $585
Mathematics
1 answer:
Elina [12.6K]3 years ago
4 0

answer
C.

reason
First you must divide, 650/10. Which will equal to 65. Subtract 65 from 650. You will end up with 585. To represent the amount that is left would be 9/10,
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Determine the volume in litres of 14kg of petrol of density 700kg/m³​
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p =  \frac{m}{v}

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In a completely randomized experimental design, three brands of paper towels were tested for their ability to absorb water. Equa
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Yes. At this significance level, there is evidence to support the claim that there is a difference in the ability of the brands to absorb water.

Step-by-step explanation:

<em>The question is incomplete:</em>

<em>The significance level is 0.05.</em>

<em>The data is:</em>

<em>Brand X: 91, 100, 88, 89</em>

<em>Brand Y: 99, 96, 94, 99</em>

<em>Brand Z: 83, 88, 89, 76</em>

We have to check if there is a significant difference between the absorbency rating of each brand.

Null hypothesis: all means are equal

H_0:\mu_x=\mu_y=\mu_z

Alternative hypothesis: the means are not equal

H_a: \mu_x\neq\mu_y\neq\mu_z

We have to apply a one-way ANOVA

We start by calculating the standard deviation for each brand:

s_x^2=30,\,\,s_y^2=6,\,\,s_z^2=35.33

Then, we calculate the mean standard error (MSE):

MSE=(\sum s_i^2)/a=(30+6+35.33)/3=71.33/3=23.78

Now, we calculate the mean square between (MSB), but we previously have to know the sample means and the mean of the sample means:

M_x=92,\,\,M_y=97,\,\,M_z=84\\\\M=(92+97+84)/3=91

The MSB is then:

s^2=\dfrac{\sum(M_i-M)^2}{N-1}\\\\\\s^2=\dfrac{(92-91)^2+(97-91)^2+(84-91)^2}{3-1}\\\\\\s^2=\dfrac{1+36+49}{2}=\dfrac{86}{2}=43\\\\\\\\MSB=ns^2=4*43=172

Now we calculate the F statistic as:

F=MSB/MSE=172/23.78=7.23

The degrees of freedom of the numerator are:

dfn=a-1=3-1=2

The degrees of freedom of the denominator are:

dfd=N-a=3*4-3=12-3=9

The P-value of F=7.23, dfn=2 and dfd=9 is:

P-value=P(F>7.23)=0.01342

As the P-value (0.013) is smaller than the significance level (0.05), the null hypothesis is rejected.

There is evidence to support the claim that there is a difference in the ability of the brands to absorb water.

3 0
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