Andrew charges $18 dollars ofr each lawn he mows. Suppose he mows 17 lawns per month. How much money will Andrew make per month?

Find the area of the kite with diagonals a= 12 and b=9

Gnoma [55]

maybe 108 is the answer.

4 0

2 years ago

Lara went to the nursery and spent $140 she purchased 4 plants each of the plant costs the same price what is the cost of 1 plan

trapecia [35]

Answer:

$35 per plant

Step-by-step explanation:

You take the 140 and divide that by 4 to get your answer.

8 0

3 years ago

Solve the equation. 5/7=10/x+2

skelet666 [1.2K]

Answer:

$x=-\frac{70}{9}$

Step-by-step explanation:

Isolate the variable by doing operations to both sides of the equation.

The work is shown below:

$\frac{5}{7} =\frac{10}{x} +2\\\\\frac{5}{7} - 2=\frac{10}{x} +2 - 2\\\\-\frac{9}{7} = \frac{10}{x}\\\\ -\frac{9}{7}*x = \frac{10}{x}*x\\\\ -\frac{9}{7}x = 10\\\\ -\frac{9}{7}x * (-\frac{7}{9})= 10 * (-\frac{7}{9}) \\\\x=-\frac{70}{9}$

6 0

3 years ago

Read 2 more answers

A local car dealer claims that 25% of all cars in San Francisco are blue. You take a random sample of 600 cars in San Francisco

sammy [17]

Answer:

No, we can't reject the dealer's claim with a significance level of 0.05.

Step-by-step explanation:

We are given that a local car dealer claims that 25% of all cars in San Francisco are blue.

You take a random sample of 600 cars in San Francisco and find that 141 are blue.

Let p = proportion of all cars in San Francisco who are blue

SO, Null Hypothesis, $H_0$ : p = 25% {means that 25% of all cars in San Francisco are blue}

Alternate Hypothesis, $H_A$ : p $\neq$ 25% {means that % of all cars in San Francisco who are blue is different from 25%}

The test statistics that will be used here is One-sample z proportion statistics;

T.S. = $\frac{\hat p-p}{{\sqrt{\frac{\hat p(1-\hat p)}{n} } } } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of 600 cars in San Francisco who are blue = $\frac{141}{600}$ = 0.235

n = sample of cars = 600

So, test statistics = $\frac{0.235-0.25}{{\sqrt{\frac{0.235(1-0.235)}{600} } } } }$

= -0.866

Now at 0.05 significance level, the z table gives critical values of -1.96 and 1.96 for two-tailed test. Since our test statistics lies within the range of critical values of z so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that 25% of all cars in San Francisco are blue which means the dealer's claim was correct.

4 0

3 years ago

Which expression is equivalent to (4x^(3)y^(5))(3x^(5)y)^(2)

marshall27 [118]

Answer:

$(4x^3y^5)(3x^5y)^2 = 36*x^{13}y^7$