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3 years ago

Can somebody help me with this question

Mathematics

2 answers:

frosja888 [35]3 years ago

8 0

A solution is when the two lines intersect, when they have the same (x,y) point. So we can say y=y which is the same as saying:

2x-15=5x

-3x-15=0

-3x=15

x=-5, now use either to solve for the y coordinate...y=5x becomes:

y=5(-5)=-25 so the solution is the point:

(-5, -25)

Alexus [3.1K]3 years ago

3 0

We start with:

y=2x-15
y=5x

Now, we have a perfectly set-up variable we can get rid of: y.

But we need to make either the top or bottom equation negative, so for simplicity, let's make the bottom one negative.

y=2x-15
(-y)=(-5x)

Now simplify.

-3x-15=0

Add 3x on both sides.

3x=15

Divide by 3 on both sides.

x=5

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3 years ago

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siniylev [52]

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3 years ago

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ludmilkaskok [199]

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Step-by-step explanation:

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3 years ago

A new dental bondlng agent. When bonding teeth, orthodontists must maintain a dry field. A new bonding adhesive(called "Smartbon

Gemiola [76]

Answer:

(a) H₀: μ ≥ 5.70 vs. Hₐ:μ < 5.70

(b) The rejection region is (t₀.₀₁,₉ ≤ -2.821).

(d) The true mean breaking strength of the new bonding adhesive is less than 5.70 Mpa.

Step-by-step explanation:

A hypothesis test should be conducted to determine that the if the true mean breaking strength of the new bonding adhesive is less than 5.70 Mpa.

(a)

The hypothesis is:

H₀: The true mean breaking strength of the new bonding adhesive is not less than 5.70 Mpa, i.e. μ ≥ 5.70.

Hₐ: The true mean breaking strength of the new bonding adhesive is less than 5.70 Mpa, i.e. μ < 5.70.

(b)

The alternate hypothesis indicates that the hypothesis test is left-tailed.

The rejection region for the left tailed test will be towards the lower tail of the t-distribution curve.

The significance level of the test is: α = 0.01.

The critical value is:

$t_{\alpha ,(n-1)}=t_{0.01,(10-1)}=t_{0.01,9}$

Use the t-table for the critical value.

$t_{\alpha ,(n-1)}=t_{0.01,9}=-2.821$

Since rejection region is in the lower tail the critical value will be negative.

Thus, the rejection region is (t₀.₀₁,₉ ≤ -2.821).

(c)

The test statistic value is:

$t=\frac{\bar x-\mu}{s/\sqrt{n}}$

Given:

$\bar x=5.07\\s=0.46\\n=10\\\mu=5.70$

Compute the value of the t-statistic as follows:

$t=\frac{\bar x-\mu}{s/\sqrt{n}}=\frac{5.07-5.70}{0.46/\sqrt{10}} =-4.33$

The value of the test statistic is -4.33.

(d)

The value of the test is less than the critical value.

$t=-4.33$

This implies that the test statistic lies in the rejection region.

Hence the null hypothesis will be rejected at 1% significance level.

Conclusion:

As the null hypothesis is rejected it can be concluded that the true mean breaking strength of the new bonding adhesive is less than 5.70 Mpa.

(e)

The conditions required for the t-test for single mean to be valid is:

The data should be continuous.
The parent population should be normally distributed.
The sample should be randomly selected.

5 0

3 years ago

The College Board SAT college entrance exam consists of three parts: math, writing and critical reading (The World Almanac 2012)

Wittaler [7]

Answer:

Yes, there is a difference between the population mean for the math scores and the population mean for the writing scores.

Test Statistics = $\frac{Dbar - \mu_D}{\frac{s_D}{\sqrt{n} } }$ follows $t_n_- _1$ .

Step-by-step explanation:

We are provided with the sample data showing the math and writing scores for a sample of twelve students who took the SAT ;

Let A = Math Scores ,B = Writing Scores and D = difference between both

So, $\mu_A$ = Population mean for the math scores

$\mu_B$ = Population mean for the writing scores

Let $\mu_D$ = Difference between the population mean for the math scores and the population mean for the writing scores.

Null Hypothesis, $H_0$ : $\mu_A = \mu_B$ or $\mu_D$ = 0

Alternate Hypothesis, $H_1$ : $\mu_A \neq \mu_B$ or $\mu_D \neq$ 0

Hence, Test Statistics used here will be;

$\frac{Dbar - \mu_D}{\frac{s_D}{\sqrt{n} } }$ follows $t_n_- _1$ where, Dbar = Bbar - Abar

$s_D$ = $\sqrt{\frac{\sum D_i^{2}-n*(Dbar)^{2}}{n-1}}$

n = 12

Student Math scores (A) Writing scores (B) D = B - A

1 540 474 -66

2 432 380 -52

3 528 463 -65

4 574 612 38

5 448 420 -28

6 502 526 24

7 480 430 -50

8 499 459 -40

9 610 615 5

10 572 541 -31

11 390 335 -55

12 593 613 20

Now Dbar = Bbar - Abar = 489 - 514 = -25

Bbar = $\frac{\sum B_i}{n}$ = $\frac{474+380+463+612+420+526+430+459+615+541+335+613}{12}$ = 489

Abar = $\frac{\sum A_i}{n}$ = $\frac{540+432+528+574+448+502+480+499+610+572+390+593}{12}$ = 514

∑ $D_i^{2}$ = 22600 and $s_D$ = $\sqrt{\frac{\sum D_i^{2}-n*(Dbar)^{2}}{n-1}}$ = $\sqrt{\frac{22600 - 12*(-25)^{2} }{12-1} }$ = 37.05

So, Test statistics = $\frac{Dbar - \mu_D}{\frac{s_D}{\sqrt{n} } }$ follows $t_n_- _1$

= $\frac{-25 - 0}{\frac{37.05}{\sqrt{12} } }$ follows $t_1_1$ = -2.34

Now at 5% level of significance our t table is giving critical values of -2.201 and 2.201 for two tail test. Since our test statistics doesn't fall between these two values as it is less than -2.201 so we have sufficient evidence to reject null hypothesis as our test statistics fall in the rejection region .

Therefore, we conclude that there is a difference between the population mean for the math scores and the population mean for the writing scores.

8 0

3 years ago