Name three equal ratios from the graph

Rule: add 1, multiply by 2, first term:2

lesya692 [45]

2+1=3x2=6

Sooo, your answer would be 6

4 0

3 years ago

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Given the expression 12 over 12 to the third power.

4vir4ik [10]

Part 1:
Before we begin, you need to remember the following rule:
$\frac{x^a}{x^b} = x^{a-b}$

The given expression is:
$\frac{12}{12^3}$

Since the base is the same in both numerator and denominator, we can apply the above rule. This means that:
$\frac{12}{12^3}$ = 12¹⁻³ = 12⁻²

Part 2:
Before we begin, you need to remember the following rule:
x⁻ᵃ = $\frac{1}{x^a}$

Now, from part 1, we simplified the expression into 12⁻²
Since the power is negative, we can apply the above rule.
This means that:
12⁻² = $\frac{1}{(12)^2} = \frac{1}{12*12} = \frac{1}{144}$

Hope this helps :)

5 0

3 years ago

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Identify the scale factor when we scale up from Figure S to T. Just write the number.

Sergeu [11.5K]

Answer:

Scale factor = 4

Step-by-step explanation:

Scale factor = $\frac{\text{Dimension of the image}}{\text{Dimension of preimage or original}}$

Since, S is being dilated with a scale factor to form T,

So, the figure S is the preimage and figure T will be the image,

Length of one side of image T = 8

Length of one side of preimage = 2

Therefore, Scale factor = $\frac{8}{2}$

= 4

Scale factor by which figure S has been dilated to form figure T is 4.

3 0

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).

(c) The value of the test statistic is -4.33.

(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

ABCD=STQR. What is the measurement of CD?

USPshnik [31]

Step-by-step explanation:

$\because \: ABCD=STQR...(Given) \\ \\ \therefore \: CD = QR \\ \\ \because \: QR = 21 \: in \\ \\ \huge \red{ \boxed{\therefore \: CD =21 \: in }}$

6 0

3 years ago

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