How many students in ms. Leal class are more than 42 inches tall?

Given 2x+7=27 and 3x+1=28, does 2x+7=3x+1? Explain

serious [3.7K]

No, in 2x+7=27, x=10. In 3x+1=28 x =9, so it doesn't.

7 0

3 years ago

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H(2) = 3.c +3<br> g(x) = -4x+1<br> Find (h+g) (10)

choli [55]

Good g bxbmxcgxcy sc tftrgyt tzj scandal

7 0

3 years ago

Hendrick wants to enlarge a photo that is 4 inches wide and 6 inches tall. The enlarged photo keeps the same ratio. How y'all is

zhuklara [117]

Answer:

Step-by-step explanation:

Alright, lets get started.

The original photo size is 4 inches wide and 6 inches tall.

So, the ratio of width and height will be = $\frac{4}{6}=\frac{2}{3}$

The new enlarged photo will be of the same ratio means 2:3

The width of enlarged photo is given as 12 inches.

Suppose new height of enlarged photo is H, so

$\frac{12}{H}=\frac{2}{3}$

Cross multiplying

$2H=36$

Dividing 2 in both sides

$H=18$ inches

So the height of new enlagred photo will be 18 inches. : Answer

Hope it will help :)

7 0

3 years ago

Simplify the expression.

OlgaM077 [116]

Answer:

5.1

Step-by-step explanation:

-44.288-31.6 / -3.1(6-1.2)

Add -44.288 and -31.6

= -75.888 / -3.1(6-1.2)

Subtract 1.2 from 6

= -75.888 / -3.1(4.8)

Multiply -3.1 and 4.8

= -75.888 / -14.88

Divide -75.888 by -14.88

-75.888 / -14.88

=5.1

6 0

3 years ago

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The following 5 questions are based on this information: An economist claims that average weekly food expenditure of households

liq [111]

Answer:

Null hypothesis: $\mu_{1}-\mu_{0}\leq 0$

Alternative hypothesis: $\mu_{1}-\mu_{2}>0$

$SE=\sqrt{\frac{12.5^2}{35}+\frac{9.25^2}{30}}=2.705$

b) 2.70

$t=\frac{(164-159)-0}{\sqrt{\frac{12.5^2}{35}+\frac{9.25^2}{30}}}=1.850$

b. 1.85

$p_v =P(Z>1.85)=0.032$

b. 0.03

a. We can reject H(0) in favor of H(a)

Step-by-step explanation:

Data given and notation

$\bar X_{1}=164$ represent the mean for the sample 1

$\bar X_{2}=159$ represent the mean for the sample 2

$\sigma_{1}=12.5$ represent the population standard deviation for the sample 1

$s_{2}=9.25$ represent the population standard deviation for the sample B2

$n_{1}=35$ sample size selected 1

$n_{2}=30$ sample size selected 2

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the expenditure of households in City 1 is more than that of households in City 2, the system of hypothesis would be:

Null hypothesis: $\mu_{1}-\mu_{0}\leq 0$

Alternative hypothesis: $\mu_{1}-\mu_{2}>0$

We know the population deviations, so for this case is better apply a z test to compare means, and the statistic is given by:

$z=\frac{(\bar X_{1}-\bar X_{2})-0}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

Standard error

The standard error on this case is given by:

$SE=\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}$

Replacing the values that we have we got:

$SE=\sqrt{\frac{12.5^2}{35}+\frac{9.25^2}{30}}=2.705$

b. 2.70

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{(164-159)-0}{\sqrt{\frac{12.5^2}{35}+\frac{9.25^2}{30}}}=1.850$

P-value

Since is a one side right tailed test the p value would be:

$p_v =P(Z>1.85)=0.032$

b. 0.03

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis.

a. We can reject H(0) in favor of H(a)

4 0

3 years ago