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3 years ago

What is the imgae point of (-2,-3) after a translation right 2 units and 4 up units?

Mathematics

1 answer:

kherson [118]3 years ago

5 0

the image point of (-2,-3) after a translation right 2 units and 4 up units is (0,1) .

<u>Step-by-step explanation:</u>

Here we have , to find the image point of (-2,-3) after a translation right 2 units and 4 up units . Let's find out:

Initially we have the point as (-2,-3) , Following transformations are done :

a translation right 2 units :

The point is (-2,-3) , let $(x,y)=(-2,-3)$ . Translation is 2 units right , means there is change in x coordinate by +2 , i.e.

⇒ $(x,y)=(-2,-3)$

⇒ $(x+2,y)=(-2+2,-3)$

⇒ $(x+2,y)=(0,-3)$

a translation 4 up units:

The point is (0,-3) , let $(x,y)=(0,-3)$ . Translation is 4 units up , means there is change in y coordinate by +4 , i.e.

⇒ $(x,y)=(0,-3)$

⇒ $(x,y+4)=(0,-3+4)$

⇒ $(x,y+4)=(0,1)$

Therefore , the image point of (-2,-3) after a translation right 2 units and 4 up units is (0,1) .

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A random sample of 20 students yielded a mean of ¯x = 72 and a variance of s2 = 16 for scores on a college placement test in mat

Helen [10]

Answer:

The 98% confidence interval for the variance in the pounds of impurities would be $8.400 \leq \sigma^2 \leq 39.827$ .

Step-by-step explanation:

1) Data given and notation

$s^2 =16$ represent the sample variance

s=4 represent the sample standard deviation

$\bar x$ represent the sample mean

n=20 the sample size

Confidence=98% or 0.98

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population mean or variance lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

The Chi square distribution is the distribution of the sum of squared standard normal deviates .

2) Calculating the confidence interval

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

On this case the sample variance is given and for the sample deviation is just the square root of the sample variance.

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

$df=n-1=20-1=19$

Since the Confidence is 0.98 or 98%, the value of $\alpha=0.02$ and $\alpha/2 =0.01$ , and we can use excel, a calculator or a table to find the critical values.

The excel commands would be: "=CHISQ.INV(0.01,19)" "=CHISQ.INV(0.99,19)". so for this case the critical values are:

$\chi^2_{\alpha/2}=36.191$

$\chi^2_{1- \alpha/2}=7.633$

And replacing into the formula for the interval we got:

$\frac{(19)(16)}{36.191} \leq \sigma^2 \leq \frac{(19)(16)}{7.633}$

$8.400 \leq \sigma^2 \leq 39.827$

So the 98% confidence interval for the variance in the pounds of impurities would be $8.400 \leq \sigma^2 \leq 39.827$ .

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3 years ago

A liquid at temperature 7575F is placed in an oven at temperature 450450. The temperature of the liquid increases at a rate 77 t

hichkok12 [17]

Answer: $\dfrac{dT(t)}{dt}=77\times (450-T)$

Step-by-step explanation:

Given

The temperature of the liquid is $75^{\circ}F$ placed in an oven with temperature of $450^{\circ}F$ .

Initially difference in temperature of the two

$\Delta T=450-75\\\Rightarrow \Delta T=375^{\circ}F$

According to the question

$\Rightarrow \dfrac{dT(t)}{dt}=77\cdot \Delta T\\\\\Rightarrow \dfrac{dT(t)}{dt}=77\times (450-T)\quad [\text{T=75}^{\circ}F\ \text{at t=0}]$

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2 years ago

The Mean Corporation's statisticians would like to construct a hypothesis test for the mean annual profit (μ) earned by health-d

Jobisdone [24]

Missing Part of Question

Juice that is opened. It is known that the annual profits earned by health-drink franchises has a population standard deviation of $8,700

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Answer:

z = 1.157

Step-by-step explanation:

Given

Annual profit as calculated = $90,300.

H0: μ = 89,000

Ha: μ > 89,000

σ = 8700

n = 60

Calculate the test statistic (z) that corresponds to the sample and hypotheses.

Test statistic (z) is calculated by

z = (x - μ)/(σ/√n)

Where x = 90,300

μ = 89,000

σ = 8700

n = 60

z = (90,300 - 89,000)/(8700/√60)

z = 1.157443298866584

z = 1.157 ----- Approximated

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3 years ago

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Vsevolod [243]

I think u should try (t-23) because -8 minus - 15 is = to 23 so yeah add the t in front of it

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3 years ago

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Brums [2.3K]

The question is missing the figure which is attached below.

Answer:

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Step-by-step explanation:

Given:

The two prisms are similar.

Volume of the smaller prism (V₁) = 60 cm³

Length of the smaller prism (l₁) = 5 cm

Length of the larger prism (l₂) = 15 cm

Now, we know that, for similar figures, the dimensions of the figure are in proportion to each other. Therefore,

$\frac{l_2}{l_1}=\frac{b_2}{b_1}=\frac{h_2}{h_1}=k(constant)\\\\Where,\\b_1,b_2\to\ widths\ of smaller\ and\ larger\ prisms\ respectively\\\\h_1,h_2\to\ heights\ of\ smaller\ and\ larger\ prisms\ respectively$