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2 years ago

What is the missing number?

Mathematics

2 answers:

Lynna [10]2 years ago

8 0

B. 45. Five goes into 15 three times, and 5 goes into 225 45 times.

rjkz [21]2 years ago

5 0

It would be 45 because 15 divided by 5 is 3. Then you do the same for 225. 225/5 is 45

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I really don't understand help

Zielflug [23.3K]

Answer:

128 ft sq.

Step-by-step explanation:

The way you find the area of a square is by multiplying one side by another, and since it is a square, the sides are the same. This means that if we have a square with a side length of x, the way to find the area of that square is to multiply x by x, which can be written as $x^2$ . Now, since we're doubling this, we can say that double Mary's garden is $2 * x^2$ . If we substitute 8 in for x, we have to multiply $8^2$ by 2 or 8 by 8 by 2 to find our answer. 8 times 8 is 64 and 64 times 2 is 128.

5 0

2 years ago

What is the answer??

Lemur [1.5K]

Answer:

h(3) = - 140

Step-by-step explanation:

Generate the terms in the sequence by substituting n = 2 and 3 into h(n)

h(2) = h(2 - 1) × 2 = h(1) × 2 = - 35 × 2 = - 70

h(3) = h(3 - 1) × 2 = h(2) × 2 = - 70 × 2 = - 140

8 0

3 years ago

use an algebraic equation to solve the problem. the sides of a triangle are in the ratio 3:4:5. what is the length of each side

STatiana [176]

This should be the equation:

n + 4/3n + 5/3n = 90

5 0

3 years ago

Put into slope intercept form

Rudiy27

The slope is 2 units down for 9 units over, so is -2/9. The y-intercept is 4.

In slope-intercept form the equation of the line is
y = (-2/9)x +4

4 0

3 years ago

In a study comparing various methods of gold plating, 7 printed circuit edge connectors were gold-plated with control-immersion

S_A_V [24]

Answer:

99% confidence interval for the difference between the mean thicknesses produced by the two methods is [0.099 μm , 0.901 μm].

Step-by-step explanation:

We are given that in a study comparing various methods of gold plating, 7 printed circuit edge connectors were gold-plated with control-immersion tip plating. The average gold thickness was 1.5 μm, with a standard deviation of 0.25 μm.

Five connectors were masked and then plated with total immersion plating. The average gold thickness was 1.0 μm, with a standard deviation of 0.15 μm.

Firstly, the pivotal quantity for 99% confidence interval for the difference between the population mean is given by;

P.Q. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t__n__1+_n__2-2$

where, $\bar X_1$ = average gold thickness of control-immersion tip plating = 1.5 μm

$\bar X_2$ = average gold thickness of total immersion plating = 1.0 μm

$s_1$ = sample standard deviation of control-immersion tip plating = 0.25 μm

$s_2$ = sample standard deviation of total immersion plating = 0.15 μm

$n_1$ = sample of printed circuit edge connectors plated with control-immersion tip plating = 7

$n_2$ = sample of connectors plated with total immersion plating = 5

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(7-1)\times 0.25^{2}+(5-1)\times 0.15^{2} }{7+5-2} }$ = 0.216

<em>Here for constructing 99% confidence interval we have used Two-sample t test statistics as we don't know about population standard deviations.</em>

So, 99% confidence interval for the difference between the mean population mean, ( $\mu_1-\mu_2$ ) is ;

P(-3.169 < $t_1_0$ < 3.169) = 0.99 {As the critical value of t at 10 degree of

freedom are -3.169 & 3.169 with P = 0.5%}

P(-3.169 < $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ < 3.169) = 0.99

P( $-3.169 \times {s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ < ${(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}$ < $3.169 \times {s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ) = 0.99

P( $(\bar X_1-\bar X_2)-3.169 \times {s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ < ( $\mu_1-\mu_2$ ) < $(\bar X_1-\bar X_2)+3.169 \times {s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ) = 0.99

<u>99% confidence interval for</u> ( $\mu_1-\mu_2$ ) =

[ $(\bar X_1-\bar X_2)-3.169 \times {s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ , $(\bar X_1-\bar X_2)+3.169 \times {s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ]

= [ $(1.5-1.0)-3.169 \times {0.216\sqrt{\frac{1}{7}+\frac{1}{5} } }$ , $(1.5-1.0)+3.169 \times {0.216\sqrt{\frac{1}{7}+\frac{1}{5} } }$ ]

= [0.099 μm , 0.901 μm]

Therefore, 99% confidence interval for the difference between the mean thicknesses produced by the two methods is [0.099 μm , 0.901 μm].

6 0

3 years ago

Read 2 more answers