Alexa knock down 70 bowling pins in 10 frames. in each frame, alexa nocked down the same number of pins. how many pins did Alexa knock down in each frame?

Answer:

Cost of each pillow is $120.

Step-by-step explanation:

Let P denotes cost of each pillow and M denotes cost of mattress.

Then, given cost of mattress is $769

also, Ally gave the cashier $1300 and received $51 change.

Total amount Ally paid = total amount she gave - total amount she receives back.

                                     = $( 1300 - 51 )

                                     = $1249

Since, Ally brought one mattress and 4 pillow . Thus, Total cost is given as,

⇒ M + 4 P = 1249

⇒ 769 + 4 P = 1249

⇒ 4 P = 1249 - 769

⇒ 4 P = 480

⇒ P = 120

Thus, cost of each pillow is $120.

Answer:

16 dimes and 18 quarters!!

0.5x + 0.1x = 0.25x + 0.2x + 1.20

0.6x = 0.45x + 1.20

0.6x - 0.45x = 1.20

0.15x = 1.20

x = 1.20/0.15

x = 8 quarters

2x = 16 dimes

hope this helped :D

The product of matrices[tex]\(B\) and \(A\) is \([12 \, 24 \, 4 \, -2]\).[/tex]

To find the product (BA), we multiply each row of (B) by each column of (A) and sum the products.

Let's calculate the elements of the resulting matrix (BA):

  a. First row, first column:

[tex]\[ (2)(3) + (8)(0) + (0.6)(2) + (3)(-1) = 6 + 0 + 1.2 - 3 = 4.2 \][/tex]

  b. First row, second column:

[tex]\[ (2)(3) + (8)(0) + (0.6)(2) + (3)(-1) = 6 + 0 + 1.2 - 3 = 4.2 \][/tex]

  c. First row, third column:

[tex]\[ (2)(3) + (8)(0) + (0.6)(2) + (3)(-1) = 6 + 0 + 1.2 - 3 = 4.2 \][/tex]

  d. First row, fourth column:

[tex]\[ (2)(3) + (8)(0) + (0.6)(2) + (3)(-1) = 6 + 0 + 1.2 - 3 = 4.2 \][/tex]

Therefore, the first row of [tex]\(BA\) is \([4.2 \, 4.2 \, 4.2 \, 4.2]\).[/tex]

Similarly, for the second row of (BA):

  a. Second row, first column:

[tex]\[ (2)(3) + (8)(0) + (0.6)(2) + (3)(-1) = 6 + 0 + 1.2 - 3 = 4.2 \][/tex]

  b. Second row, second column:

[tex]\[ (2)(3) + (8)(0) + (0.6)(2) + (3)(-1) = 6 + 0 + 1.2 - 3 = 4.2 \][/tex]

  c. Second row, third column:

[tex]\[ (2)(3) + (8)(0) + (0.6)(2) + (3)(-1) = 6 + 0 + 1.2 - 3 = 4.2 \][/tex]

  d. Second row, fourth column:

[tex]\[ (2)(3) + (8)(0) + (0.6)(2) + (3)(-1) = 6 + 0 + 1.2 - 3 = 4.2 \][/tex]

Therefore, the second row of [tex]\(BA\) is \([4.2 \, 4.2 \, 4.2 \, 4.2]\).[/tex]

Hence, the product [tex]\(BA\) is \([12 \, 24 \, 4 \, -2]\).[/tex]

Answer:

4, -32

2, -15

Step-by-step explanation:

Trust me bro.

The Distributive Property allows you to multiply the terms within (z−5)(z+3) to obtain z² + 3z - 5z - 15. After combining like terms, the final result is z² - 2z - 15.

To use the Distributive Property to find the product of (z−5)(z+3), you multiply each term in the first parenthesis by each term in the second parenthesis. Here are the steps:

Multiply z by z, which is z².

Multiply z by +3, which gives you +3z.

Multiply -5 by z, which results in -5z.

Finally, multiply -5 by +3, which is -15.

After multiplication, you combine like terms:

z² + 3z - 5z - 15

Combine +3z and -5z to get -2z

So, the final result is z² - 2z - 15.

The average human heart beats approximately 108,000 times in one day, 39 million times in one year, and nearly 3 billion times during a 75-year lifespan. In a lifetime, your heart will beat an estimated 2.925 x 10^9 times.

The average human heart beats approximately 108,000 times in one day, 39 million times in one year, and nearly 3 billion times during a 75-year lifespan. To estimate the number of times your heart will beat in your lifetime, we can multiply the number of heartbeats in one year by the average life span:

39 million heartbeats/year x 75 years = 2.925 billion heartbeats in a lifetime

In scientific notation, this can be expressed as 2.925 x 10^9 heartbeats.

The average heart beats approximately 3 billion times during a 75-year lifespan.

The average heart beats approximately 108,000 times in one day, more than 39 million times in one year, and nearly 3 billion times during a 75-year lifespan. To estimate the number of times your heart will beat in your lifetime, you multiply the average heartbeats per year by the number of years in your lifespan.

For example, if we assume an average lifespan of 75 years, we can estimate the number of heartbeats in a lifetime:

39,000,000 heartbeats/year x 75 years = 2,925,000,000 heartbeats in a lifetime (2.9 x 10^9)

The 95% confidence interval for the population mean ( μ) is approximately (8.46,11.54), calculated from a sample of 26 elements with a mean of 10 and a standard deviation of 4.

To calculate the 95% confidence interval for the population mean (μ), we can use the formula:

Confidence interval=Sample mean±(Critical value× Sample size/Standard deviation​ )

Given:

Sample mean (xˉ ) = 10

Standard deviation (σ) = 4

Sample size (n) = 26

Confidence level = 95%

Step 1: Find the critical value from the Z-table for a 95% confidence level.

Since it's a two-tailed test, we'll find the Z-value corresponding to a cumulative probability of 0.975.

Z α/2​ =1.96

Step 2: Plug the values into the formula:

Confidence interval=10±(1.96× 26​/4​ )

Step 3: Calculate the margin of error:

Margin of error≈1.96×45.099

Margin of error≈1.96× 5.0994

Margin of error≈1.96×0.785

Margin of error≈1.5376

Step 4: Calculate the confidence interval:

Lower limit=10−1.5376

Lower limit≈8.4624

Upper limit=10+1.5376

Upper limit≈11.5376

So, the 95% confidence interval for

μ is approximately (8.4624,11.5376).

The general solution to the differential equation z″+8z′=0 is z(t) = Ae^(-8t) + B, where A and B are arbitrary constants.

The differential equation z″+8z′=0 can be solved by separating variables and integrating.

Let v = z′ be the derivative of z.

Then the equation becomes v′+8v=0.

This is a first-order linear homogeneous differential equation, which can be solved by multiplying both sides of the equation by the integrating factor e^(∫8 dt) = e^(8t).

After solving for v, we can integrate it again to find z(t) by substituting back the expression for v in terms of z.

The general solution to the differential equation is z(t) = Ae^(-8t) + Be^(0t) = Ae^(-8t) + B, where A and B are arbitrary constants.

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The water level in a paper cup shaped like a cone with radius 3 cm and height 10 cm, filled at a rate of 2 cm³/s, rises at approximately 0.025 cm per second when the water is 5 cm deep.

This question involves related rates, a concept in Calculus. We know that the volume, V, of a cone with a radius r and height h is given by the formula V = (1/3)πr²h. Given that the shape of the cup is conical, the radius and the height of the water in the cup are proportional, so we can express r as r=3h/10.

Thus, we can rewrite the volume formula in terms of h: V = 1/3 * π * (3h/10)² * h = πh³/100. Differentiating both sides with respect to time t, we get dV/dt = πh² dh/dt. We want to find dh/dt (the rate at which the water level rises) when h=5 cm and given that dV/dt (the rate at which water is poured into the cup) is 2 cm³/s.

Plugging these values into the differentiated formula, we get: 2 = π(5)² * dh/dt. Solving for dh/dt, we find that dh/dt = 2/(25π) or about 0.025 cm/s.  So, the water level is rising at a rate of approximately 0.025 cm per second when the water is 5 cm deep.

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Answer:  Our factorized term will be

2xy(2x+5)(2x-3)

Explanation:

Since we have given that

[tex]8x^3y-8x^2y-30xy[/tex]

All we need to do is to factorise this expression.

Here are the steps given below:

[tex]2xy(4x^2-4x-15)[/tex]

By manipulating the terms, we get:

[tex]2xy[(2x)^2-4x+1-16]\\\\2xy[(2x-1)^2-16]\\\\2xy[(2x-1)^2-(4)^2]\\\\2xy[(2x-1-4)(2x-1+4)]\\\\2xy[(2x-5)(2x+3)][/tex]

Alternatively,

By using splitting the middle terms,

[tex]2xy(4x^2-4x-15)\\\\2xy(4x^2-10x+6x-15)\\\\2xy[2x(2x-5)+3(2x-5)]\\\\2xy(2x-3)(2x+5)[/tex]

Hence, our factorized term will be

2xy(2x+5)(2x-3)

Answer:

Let X the random variable that represent the mean finish time for a yearly amateur auto race a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(185.19,0.341)[/tex]  

Where [tex]\mu=185.19[/tex] and [tex]\sigma=0.341[/tex]

The z score is given by this formula:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

And for a time of 184.14 we have the following z score:

[tex] z = \frac{184.14-185.19}{0.341}= -3.08[/tex]

Let Y the random variable that represent the mean finish time for the previous year auto race a population, and for this case we know the distribution for X is given by:

[tex]Y \sim N(110.4,0.137)[/tex]  

Where [tex]\mu=110.4[/tex] and [tex]\sigma=0.137[/tex]

The z score is given by this formula:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

And for a time of 110.05 we have the following z score:

[tex] z = \frac{110.05-110.4}{0.137}=-2.557[/tex]

As we can see we have a higher z score for the case of the previous year so then we have a more convincing victory on this case since represent a higher quantile in the normal standard distribution.

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the mean finish time for a yearly amateur auto race a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(185.19,0.341)[/tex]  

Where [tex]\mu=185.19[/tex] and [tex]\sigma=0.341[/tex]

The z score is given by this formula:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

And for a time of 184.14 we have the following z score:

[tex] z = \frac{184.14-185.19}{0.341}= -3.08[/tex]

Let Y the random variable that represent the mean finish time for the previous year auto race a population, and for this case we know the distribution for X is given by:

[tex]Y \sim N(110.4,0.137)[/tex]  

Where [tex]\mu=110.4[/tex] and [tex]\sigma=0.137[/tex]

The z score is given by this formula:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

And for a time of 110.05 we have the following z score:

[tex] z = \frac{110.05-110.4}{0.137}=-2.557[/tex]

As we can see we have a higher z score for the case of the previous year so then we have a more convincing victory on this case since represent a higher quantile in the normal standard distribution.

Answer:

wht the girl ontop said

Step-by-step explanation:

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Answer:

Step-by-step explanation:

Given: m∠H = 30°, m∠J = 50°, m∠P = 50°, m∠N = 100°

To prove: △HKJ ~ △LNP

Proof:

Step 1. m∠H = 30°, m∠J = 50°, m∠P = 50°, m∠N = 100° (Given)

Step 2. m∠H + m∠J + m∠K = 180° (Triangle angle sum theorem)

Step 3. 30° + 50° + m∠K = 180°(substitution property)

Step 4. 80° + m∠K = 180°(addition Property)

Step 5. m∠K = 100°(subtraction property of equality)

Step 6. m∠J = m∠P; m∠K = m∠N(substitution)

Step 7. ∠J ≅ ∠P; ∠K ≅ ∠N (If angles are equal then they are congruent)

Step 8. △HKJ ~ △LNP( AA similarity theorem)

Hence proved.

Thus, the missing step in 2 is (Triangle angle sum theorem)

Answer:

D) triangle angle sum theorem

Step-by-step explanation:

Just finished the test!!

Answer:

The equation of the resulting parabola is:

y=(x+7)^2

Step-by-step explanation:

We, know that the transformation of the type:

f(x+a) is a translation of the parent function f(x) to the left or right depending on the sign of the constant 'a'.

if:

a<0 then the translation is to the right.

and if a>0 then the translation is to the left.

It is given that the graph of the function,

let f(x)=y=x^2 is translated 7 units to the left.

This means that the equation of the resulting function will be:

y=(x+7)^2