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3 years ago

How do I do this problem?

Mathematics

1 answer:

DedPeter [7]3 years ago

5 0

If it's an isosceles triangle, the answer should be 87*42yd. In a right triangle, the area is (height*base)/2. if the height perfectly divides the triangle in half, there's no need to divide h*b by 2.

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If i have 22 more questions to go out of 42 what question am i on?

sergejj [24]

Answer: 20

Step-by-step explanation:

6 0

2 years ago

Plz help ill give brainlist to you

Lelu [443]

0.4% of $4,800 is $19.2, otherwise written as $19.20 in terms of money, apparently. This is based off the calculations of a calculator, though if you prefer to do the rough work, be my guest. The work for finding it on raw work is below.

(EX.) EQUATION : y = p% × x
EQUATION : y = 0.4% × 4800

To make this work, you gotta convert that percent into a decimal, by dividing the percentage by 100. 0.4 ÷ 100 = 0.004

EQUATION : y = 0.004 × 4800.

Once you multiply them, you get : 19.2 (otherwise known as $19.20)

6 0

3 years ago

Una recta con pendiente -2 ´pasa por el punto P(5, -1). La abscisa del punto Q que está en esa recta es 1. Encuentre la ordenada

oksian1 [2.3K]

Answer:

La ordenada de Q es 7.

Step-by-step explanation:

La ecuación de una recta es:

y = a*x + b

donde a es la pendiente y b es la ordenada al origen.

Para este problema, sabemos que la pendiente es -2

Entonces:

a = -2

y = -2*x + b

También sabemos que esta recta pasa por el punto (5, -1)

Esto significa que cuando evaluamos la ecuación en x = 5, y toma el valor y = -1

Si reemplazamos esos valores en la ecuación, obtenemos:

-1 = -2*5 + b

-1 = -10 + b

-1 + 10 = b

9 = b

Entonces la ecuación de la recta es:

y = -2*x + 9

Sabemos que el punto Q está en la línea, y la abscisa de este punto es 1.

Entonces este punto es Q (1, k)

donde k es la ordenada de este punto.

Porque nuestra línea también pasa por este punto, podemos concluir que cuando x = 1, y = k

Reemplazando eso en la ecuación tenemos:

k = -2*1 + 9 = -2 + 9 = 7

k = 7

La ordenada de Q es 7.

7 0

2 years ago

Find the length of side x in simplest radical form with rational denominator. the short leg is root 2 and the long leg is x we d

kirill115 [55]

Answer:

√2 + x

Step-by-step explanation:

To obtain the hypotenus, h:

Using Pythagoras rule :

h² = opposite² + Adjacent²

Opposite = √2

Adjacent= x

h² = (√2)² + x²

Take the square root of both sides

h = √2 + x

Hence, the hypotenus is √2 + x

5 0

2 years ago

J.J.Bean sells a wide variety of outdoor equipment and clothing. The company sells both through mail order and via the internet.

melamori03 [73]

Answer:

99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

Step-by-step explanation:

We are given that a random sample of 16 sales receipts for mail-order sales results in a mean sale amount of $74.50 with a standard deviation of $17.25.

A random sample of 9 sales receipts for internet sales results in a mean sale amount of $84.40 with a standard deviation of $21.25.

The pivotal quantity that will be used for constructing 99% confidence interval for true mean difference is given by;

P.Q. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t__n_1_+_n_2_-_2$

where, $\bar X_1$ = sample mean for mail-order sales = $74.50

$\bar X_2$ = sample mean for internet sales = $84.40

$s_1$ = sample standard deviation for mail-order purchases = $17.25

$s_2$ = sample standard deviation for internet purchases = $21.25

$n_1$ = sample of sales receipts for mail-order purchases = 16

$n_2$ = sample of sales receipts for internet purchases = 9

Also, $s_p =\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(16-1)\times 17.25^{2}+(9-1)\times 21.25^{2} }{16+9-2} }$ = 18.74

The true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is represented by ( $\mu_1-\mu_2$ ).

Now, 99% confidence interval for ( $\mu_1-\mu_2$ ) is given by;

= $(\bar X_1-\bar X_2) \pm t_(_\frac{\alpha}{2}_) \times s_p \times \sqrt{\frac{1}{n_1} +\frac{1}{n_2}}$

Here, the critical value of t at 0.5% level of significance and 23 degrees of freedom is given as 2.807.

= $(74.50-84.40) \pm (2.807 \times 18.74 \times \sqrt{\frac{1}{16} +\frac{1}{9}})$

= [$-31.82 , $12.02]

Hence, 99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

5 0

2 years ago