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2 years ago

The functions f and g are defined by these sets if input and output values.

1 answer:

5 0

(x,y)
x=input
y=output
example
we see
g=(1,2)
theefor
g(1)=2

a.
f(4)=1
g(1)=2
g(f(4))=2

b.
g(-2)=4
f(4)=1
f(g(-2))=1

c.
f(3)=5
g(5)=5
f(5)=0
f(g(f(3)))=0

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You purchase 2 action figures per week. You have 48 action figures, which is 75% of the total number of action figures you need

LiRa [457]

Answer:

Step-by-step explanation:

Let the total number of action figures = x

75% of x = 48

$\dfrac{75}{100}*x=48\\\\Multiply \ by \ 100\\\\75*x =48*100\\\\Divide \ by \ 75\\\\x=\dfrac{48*100}{75}\\\\\\x=16*4\\\\x= 64$

Remaining action figures need to be collected = total figures - number of action figures I have

= 64 - 48 = 16

Number of weeks = 16 ÷ 2 = 8

It will take 8 weeks to collect 16 action figures.

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2 years ago

Write 7,000+600+10+2 in standard form

Allisa [31]

6,612 is the answer because all you need to do is combine them by their place value.

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3 years ago

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Solve the system of linear equations below y=-2x+19 y=x+7

Julli [10]

A) y=-2x+19
B) y=x+7 Multiplying B) by -1
B) -y = -x -7 then adding it to A)
A) y = -2x + 19
3x = 12
x = 4
y = 11

3 0

3 years ago

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An engineer is comparing voltages for two types of batteries (K and Q) using a sample of 37 type K batteries and a sample of 58

Olegator [25]

Answer:

Hypothesis Test states that we will accept null hypothesis.

Step-by-step explanation:

We are given that an engineer is comparing voltages for two types of batteries (K and Q).

where, $\mu_1$ = true mean voltage for type K batteries.

$\mu_2$ = true mean voltage for type Q batteries.

So, Null Hypothesis, $H_0$ : $\mu_1 = \mu_2$ {mean voltage for these two types of

batteries is same}

Alternate Hypothesis, $H_1$ : $\mu_1 \neq \mu_2$ {mean voltage for these two types of

batteries is different]

The test statistics we use here will be :

$\frac{(X_1bar-X_2bar) - (\mu_1 - \mu_2) }{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ follows $t_n__1 + n_2 -2$

where, $X_1bar$ = 8.54 and $X_2bar$ = 8.69

$s_1$ = 0.225 and $s_2$ = 0.725

$n_1$ = 37 and $n_2$ = 58

$s_p = \sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2} } = \sqrt{\frac{(37-1)0.225^{2}+(58-1)0.725^{2} }{37+58-2} }$ = 0.585 Here, we use t test statistics because we know nothing about population standard deviations.

Test statistics = $\frac{(8.54-8.69) - 0 }{0.585\sqrt{\frac{1}{37}+\frac{1}{58} } }$ follows $t_9_3$

= -1.219

At 0.1 or 10% level of significance t table gives a critical value between (-1.671,-1.658) to (1.671,1.658) at 93 degree of freedom. Since our test statistics is more than the critical table value of t as -1.219 > (-1.671,-1.658) so we have insufficient evidence to reject null hypothesis.

Therefore, we conclude that mean voltage for these two types of batteries is same.

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2 years ago

1. A bakery packages 96 boxes of muffins today. The bakery sells 4 sizes of boxes of muffins. The table shows how many muffins a

Vadim26 [7]

Answer:1000

Step-by-step explanation:

Yes

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2 years ago