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2 years ago

Larry has saved 18,500 toward a down payment on a house. If he makes 3890 a month, How much can afford to spend on a house?

Mathematics

2 answers:

dem82 [27]2 years ago

6 0

Answer:

D) $65,180

Step-by-step explanation:

We are given that,

Amount of down payment for the house = $18,500

Amount of income per month = $3,890

Thus, we have,

Amount of income for the year = $3,890 × 12 = $46,680

So, the amount he can afford to spend = 46,680 + 18,500 = $65,180

Thus, the amount Larry can afford to spend is $65,180.

Svetlanka [38]2 years ago

5 0

Larry has saved 18,500 toward a down payment on a house. If he makes 3890 a month, He can afford to spend on a house of about <span>93,369. The answer is letter B.</span>

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A sociologist wishes to estimate the average number of automobile thefts in a large city per day within 2 automobiles. He wishes

anzhelika [568]

Answer:

$n=(\frac{2.58(4.2)}{2})^2 =29.35 \approx 30$

So the answer for this case would be n=30 rounded up to the nearest integer

Step-by-step explanation:

Information given

$ME = 2$ the margin of error desired

$\sigma =4.2$ standard deviation from previous studies

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

And on this case we have that ME =2 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

The critical value for 99% of confidence interval now can be founded using the normal distribution. The significance level is $\alpha=0.01$ and the critical value would be $z_{\alpha/2}=2.58$ , replacing into formula (b) we got:

$n=(\frac{2.58(4.2)}{2})^2 =29.35 \approx 30$

So the answer for this case would be n=30 rounded up to the nearest integer

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2 years ago

Automobile mechanics conduct diagnostic tests on 150 new cars of a particular make and model to determine the extent to which th

Aleks04 [339]

Answer:

99% Confidence interval: (0.185,0.375)

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 150

Number of cars that have faulty catalytic converters, x = 42

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99% Confidence interval:

$\hat{p}\pm z_{stat}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

$z_{critical}\text{ at}~\alpha_{0.01} = \pm 2.58$

Putting the values, we get:

$0.28\pm 2.58(\sqrt{\frac{0.28(1-0.28)}{150}}) = 0.28\pm 0.095\\\\=(0.185,0.375)$

The 99% confidence interval for the true proportion of new cars with faulty catalytic converters is (0.185,0.375)

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2 years ago

The graph represents the cost of a subscription to a newspaper. A coordinate plane showing Ferry Ride Cost with Number of Person

allochka39001 [22]

Answer:

The constant of variation is $1.50

Step-by-step explanation:

Given

Point 1 (1,2)

Point 2 (5,8)

Required

Constant of Variation

Though the graph would have assisted in answering the question; its unavailability doesn't mean the question cannot be solved.

Having said that,

the constant variation can be solved by calculating the gradient of the graph;

The gradient is often represented by m and is calculated as thus

$m = \frac{y_2 - y_1}{x_2 - x_1}$

Where

$(x_1, y_1) = (1,2)\\(x_2, y_2) = (5,8)$

By substituting values for x1,x2,y1 and y2; the gradient becomes

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$m = \frac{3}{2}$

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Hence, the constant of variation is $1.50

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2 years ago

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weeeeeb [17]

3a + a - 15 = 225 . your substituting b(a-15) into the first equation. <span />

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2 years ago

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alexdok [17]

Answer:

The upper bound is 97.5 cm

Step-by-step explanation:

The upper bound is given as the value that is larger than or equal to all values in a data set, for example, in the data set, {3, 6, 16, 23, 25}, an upper bound is 25, however, where the accuracy of the data is given, the upper bound can be found by the following relation

Where the number is given to the nearest 100, add and subtract half of hundred to obtain the upper bound and lower bound respectively

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2 years ago