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3 years ago

A taxi departs from Los Angeles and travels toward San Diego traveling at an average

Mathematics

2 answers:

kupik [55]3 years ago

7 0

Answer:

x = x0 - 55*t, being 'x' the taxi's distance to San Diego, 'x0' the distance between Los Angeles and San Diego, and 't' the time in hours since the taxi left Los Angeles.

Step-by-step explanation:

Let's call the distance from Los Angeles to San Diego "x0"

The taxi is driving toward San Diego, so as time increases, its distance to San Diego decreases, so we need to use its speed with a negative sign in our equation of distance.

Being "t" the time in hours starting when the taxi leaves Los Angeles, and "x" the taxi's distance from San Diego in instant "t", we can formulate the equation:

x = x0 - 55*t

Nostrana [21]3 years ago

6 0

Answer:

The distance is decreasing.

Step-by-step explanation:I just did it on imagine math

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What is one Half added to one tenth

mixas84 [53]

Answer:

1/2 x 1/10 = 1/20

Step-by-step explanation:

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4 years ago

Read 2 more answers

The true average diameter of ball bearings of a certain type is supposed to be 0.5 in. A one-sample t test will be carried out t

yulyashka [42]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Step-by-step explanation:

From the question we are told that

The population mean is $\mu = 0.5 \ in$

Generally the Null hypothesis is $H_o : \mu = 0. 5 \ in$

The Alternative hypothesis is $H_a : \mu \ne 0.5 \ in$

Considering the parameter given for part a

The sample size is n = 15

The test statistics is t = 1.66

The level of significance $\alpha = 0.05$

The degree of freedom is evaluated as

$df = n- 1$

$df = 15- 1$

$df = 14$

Using the critical value calculator at (social science statistics web site )

$t_{\frac{\alpha}{2} ,df } = t_{\frac{0.05 }{2} ,14} = 2.145$

We are making use of this $t_{\frac{\alpha }{2} }$ because it is a one-tail test

Looking at the value of t and $t_{\frac{\alpha }{2} }$ the we see that $t < t_{\frac{\alpha }{2} }$ so the null hypothesis would not be rejected

Considering the parameter given for part b

The sample size is n = 15

The test statistics is t = -1.66

The level of significance $\alpha = 0.05$

The degree of freedom is evaluated as

$df = n- 1$

$df = 15- 1$

$df = 14$

Using the critical value calculator at (social science statistics web site )

$t_{\frac{\alpha}{2} ,df } = t_{\frac{0.05 }{2} ,14} = -2.145$

Looking at the value of t and $t_{\frac{\alpha}{2} ,df }$ the we see that t does not lie in the area covered by $t_{\frac{\alpha}{2} , df }$ (i.e the area from -2.145 downwards on the normal distribution curve ) hence we fail to reject the null hypothesis

Considering the parameter given for part c

The sample size is n = 26

The test statistics is t = -2.55

The level of significance $\alpha = 0.01$

The degree of freedom is evaluated as

$df = n- 1$

$df = 26- 1$

$df = 25$

Using the critical value calculator at (social science statistics web site )

$t_{\frac{\alpha}{2} ,df } = t_{\frac{0.01 }{2} ,25} = 2.787$

Looking at the value of t and $t_{\frac{\alpha }{2} }$ the we see that t does not lie in the area covered by $t_{\alpha , df }$ (i.e the area from -2.787 downwards on the normal distribution curve ) hence we fail to reject the null hypothesis

Considering the parameter given for part d

The sample size is n = 26

The test statistics is t = -3.95

The level of significance $\alpha = 0.01$

The degree of freedom is evaluated as

$df = n- 1$

$df = 26- 1$

$df = 25$

Using the critical value calculator at (social science statistics web site )

$t_{\frac{\alpha}{2} ,df } = t_{\frac{0.01 }{2} ,25} = -2.787$

Looking at the value of t and $t_{\frac{\alpha}{2} }$ the we see that t lies in the area covered by $t_{\alpha , df }$ (i.e the area from -2.787 downwards on the normal distribution curve ) hence we reject the null hypothesis

6 0

3 years ago

If the sides of a square are increased by 20%, by what percentage does the area of the square increase?

Elena-2011 [213]

If the side of the square is a it becomes 1.2a after the increase and its area becomes 1.44a². So the increase is 1.44-1=0.44 or 44%.

3 0

3 years ago

How to do this step by step

Marina CMI [18]

1. 36 = 6·6 and 42 = 6·7 have a common factor of 6.

... x⁴ and x² have a common factor of x²

The GCF is 6x².

The factorization is 36x⁴ -42x² = 6x²(6x² -7)

2. All coefficients are multiples of 4. All variable factors are multiples of x³.

The GCF is 4x³.

The factorization is 4x⁵ -8x⁴ -4x³ = 4x³(x² -2x -1)

3. The GCF of coefficients 6 and 15 is (15 mod 6) = 3, which is also a factor of the other coefficients. The lowest power of m, which is m² is also a factor of the other terms.

The GCF is 3m².

The factorization is 6m⁵ -15m⁴ -21m³ +27m² = 3m²(2m³ -5m² -7m +9)

5 0

3 years ago

Anika buys a pair of shin guards for $8.58 and serveral pairs of soccer socks for $4.29 per pair. If she spends 25.74 before tax

Kryger [21]

Answer: 4 pairs of socks

Step-by-step explanation:

According to the described situation, Anika spends $\$25.74$ in 1 pair of shin guards for $$8.58$ and $x$ pairs of socks for $\$4.29$ each pair.