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3 years ago

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A student charges $31.50 for 3 hours of babysitting. If the amount the student earns babysitting is proportional to the hours sp

ent babysitting, which equation models the amount of dollars, y, the student earns for x hours of babysitting? Group of answer choices

1 answer:

Juliette [100K]3 years ago

3 0

First you find the proportionality: 31.50/3=10.5
So y = 10.5 x

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A local school district requires all fourth graders to complete a 20-minute test containing 200 multiple-choice mathematics prob

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The component of mathematics fluency is most likely the goal of this type of assessment is Rate.

GIven,

A local school district requires all fourth graders to complete a 20-minute test containing 200 multiple-choice mathematics problems.

The component of mathematics fluency is most likely the goal of this type of assessment is Rate.

The school district is assessing the rate because they are seeing how many questions students can answer in a given time frame.

Hence, The component of mathematics fluency is most likely the goal of this type of assessment is Rate.

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2 years ago

Write the equation of the line in Polnt-Slope form that goes through the point

marshall27 [118]

Answer:

y-5=-3(x-10)

Step-by-step explanation:

y-y1=m(x-x1)

y-5=-3(x-10)

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3 years ago

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Classify each number below as a rational number or an irrational number.

vovangra [49]

Answer:

rational

rational

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Step-by-step explanation:

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2 years ago

PLEASE HELP ME I NEED THIS ANSWER FOR ALGEBRA!

Marizza181 [45]

Answer:

$radius = 6.2$

Step-by-step explanation:

so you already have the formula, which is

$\sqrt[3]{ \frac{3v}{4\pi} }$

the v represents Volume.

and 3v would be 3×volume.

$\sqrt[3]{ \frac{3 \times 1000}{4 \times \pi} }$

$\sqrt[3]{ \frac{3000}{12.56637061} }$

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4 0

3 years ago

Randomly selected 110 student cars have ages with a mean of 8 years and a standard deviation of 3.6 years, while randomly select

monitta

Answer:

1. Yes, there is sufficient evidence to support the claim that student cars are older than faculty cars.

2. The 98% confidence interval for the difference between the two population means is [1.432 years, 3.968 years].

Step-by-step explanation:

We are given that randomly selected 110 student cars to have ages with a mean of 8 years and a standard deviation of 3.6 years, while randomly selected 75 faculty cars to have ages with a mean of 5.3 years and a standard deviation of 3.7 years.

Let $\mu_1$ = <em>mean age of student cars.</em>

$\mu_2$ = <em>mean age of faculty cars.</em>

So, Null Hypothesis, $H_0$ : $\mu_1 \leq \mu_2$ {means that the student cars are younger than or equal to faculty cars}

Alternate Hypothesis, $H_A$ : $\mu_1>\mu_2$ {means that the student cars are older than faculty cars}

(1) The test statistics that will be used here is <u>Two-sample t-test statistics</u> because we don't know about the population standard deviations;

T.S. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)} {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t_n_1_+_n_2_-_2$

where, $\bar X_1$ = sample mean age of student cars = 8 years

$\bar X_2$ = sample mean age of faculty cars = 5.3 years

$s_1$ = sample standard deviation of student cars = 3.6 years

$s_2$ = sample standard deviation of student cars = 3.7 years

$n_1$ = sample of student cars = 110

$n_2$ = sample of faculty cars = 75

Also, $s_p=\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(110-1)\times 3.6^{2}+(75-1)\times 3.7^{2} }{110+75-2} }$ = 3.641

So, <u><em>the test statistics</em></u> = $\frac{(8-5.3)-(0)} {3.641 \times \sqrt{\frac{1}{110}+\frac{1}{75} } }$ ~ $t_1_8_3$

= 4.952

The value of t-test statistics is 4.952.

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

Therefore, we support the claim that student cars are older than faculty cars.

(2) The 98% confidence interval for the difference between the two population means ( $\mu_1-\mu_2$ ) is given by;

98% C.I. for ( $\mu_1-\mu_2$ ) = $(\bar X_1-\bar X_2) \pm (t_(_\frac{\alpha}{2}_) \times s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} })$

= $(8-5.3) \pm (2.326 \times 3.641 \times \sqrt{\frac{1}{110}+\frac{1}{75} })$

= $[2.7 \pm 1.268]$

= [1.432, 3.968]

Here, the critical value of t at a 1% level of significance is 2.326.

Hence, the 98% confidence interval for the difference between the two population means is [1.432 years, 3.968 years].

7 0

3 years ago